Comprehensive Study Guide on Infinite Series: Convergence, Divergence, and Analytical Tests

Fundamental Definitions of Infinite Series

• Sum of Finite Elements:

  • Given a sequence ${a_n}$, the summation notation denotes the sum of the first $n$ elements:     $a_1 + a_2 + \dots + a_n = \sum_{k=1}^n a_k$

• Definition of an Infinite Series:

  • The sum of an infinite number of elements of the sequence ${a_n}$ in which every element is followed by its next element, denoted $a_1 + a_2 + a_3 + \dots \infty$, is called an infinite series.
  • It is formally represented as:     $\sum_{n=1}^\infty a_n \quad \text{or simply} \quad \sum a_n$
  • Sometimes an infinite series is indexed from $n = 0$:     $\sum_{n=0}^\infty a_n$     in which case the initial term is $a_0$.

• Sequence of Partial Sums:

  • Associated with the infinite series $a_1 + a_2 + \dots \infty$ is a sequence of partial sums ${S_n}$ defined by:     $S_n = a_1 + a_2 + \dots + a_n = \sum_{k=1}^n a_k$
  • $S_n$ is termed the $n^{\text{th}}$ partial sum of the series $\sum a_n$.

• Convergence, Divergence, and Oscillation:

  • Convergence: If the sequence of $n^{\text{th}}$ partial sums ${S_n}$ converges to a finite limit $S$, the series $\sum_{n=1}^\infty a_n$ is defined to be convergent. The limit $S$ is defined as the sum of the infinite series:     $\sum_{n=1}^\infty a_n = a_1 + a_2 + \dots \infty = S$
  • Divergence: The series $\sum a_n$ is said to be divergent if the sequence of partial sums ${S_n}$ diverges (i.e., $\lim_{n \to \infty} (S_n) = +\infty$ or $-\infty$).
  • Oscillation: The series $\sum a_n$ is said to be oscillatory if the sequence of partial sums ${S_n}$ oscillates (finitely or infinitely).

The Geometric Series

• Definition and Structure:

  • The geometric series is defined as:     $\sum_{n=0}^\infty r^n = 1 + r + r^2 + \dots \infty$

• Convergence and Sum ($|r| < 1$):

  • The geometric series converges if and only if $|r| < 1$.
  • The sum of the convergent series is:     $S = \frac{1}{1 - r}$
  • Derivation:
  • The $n^{\text{th}}$ partial sum of the series is:       $S_n = 1 + r + r^2 + \dots + r^{n-1} = \frac{1 - r^n}{1 - r} = \frac{1}{1 - r} - \frac{r^n}{1 - r}$
  • Taking the limit as $n \to \infty$:       $\lim_{n \to \infty} (S_n) = \frac{1}{1 - r} - \frac{1}{1 - r} \lim_{n \to \infty} (r^n)$
  • For $|r| < 1$, $\lim_{n \to \infty} (r^n) = 0$, yielding:       $\lim_{n \to \infty} (S_n) = \frac{1}{1 - r}$
  • Therefore, the sequence ${S_n}$ converges to $\frac{1}{1 - r}$, and:       $1 + r + r^2 + \dots \infty = \frac{1}{1 - r}$

• Behavior for Other Values of $r$:

  • Case $r \ge 1$:$\lim_{n \to \infty} (S_n) = +\infty$     The series diverges to $+\infty$.
  • Case $r = -1$:$\lim_{n \to \infty} (S_n) \text{ oscillates between } 0 \text{ and } 1$     The series oscillates finitely.
  • Case $r < -1$:$\lim_{n \to \infty} (S_n) \text{ oscillates between } -\infty \text{ and } +\infty$     The series oscillates infinitely.

Cauchy's General Principle of Convergence

• Theorem (1.1) Statement:

  • The series $\sum a_n$ converges if and only if for every $\epsilon > 0$, there exists a positive integer $N$ such that for all positive integers $m, n$ satisfying $n > m > N$:     $|S_n - S_m| < \epsilon$     which is equivalent to:     $|a_{m+1} + a_{m+2} + \dots + a_n| < \epsilon$

• Proof:

  • Let $S_n = a_1 + a_2 + \dots + a_n$.
  • The series $\sum a_n$ is convergent if and only if the sequence of partial sums ${S_n}$ is convergent.
  • By Cauchy's Criterion for convergence of a sequence, ${S_n}$ is convergent if and only if for any $\epsilon > 0$, there exists a positive integer $N$ such that for all $n > m > N$:     $|S_n - S_m| < \epsilon$
  • Expanding the difference of partial sums:     $S_n - S_m = (a_1 + a_2 + \dots + a_n) - (a_1 + a_2 + \dots + a_m) = a_{m+1} + a_{m+2} + \dots + a_n$
  • Hence:     $|a_{m+1} + a_{m+2} + \dots + a_n| < \epsilon$

• Alternative Formulations of the Criterion:

  • Setting $m = N + 1$:     $n > N \implies |a_{N+2} + a_{N+3} + \dots + a_n| < \epsilon$
  • Setting $m = N + 1$ and $n = N + p$ where $p > 1$:     $|a_{N+2} + a_{N+3} + \dots + a_{N+p}| < \epsilon \quad \text{for all positive integers } p > 1$

• Example-Set (1.2) — Divergence of the Harmonic Series:

  • To demonstrate using Cauchy's general principle that $\sum_{n=1}^\infty \frac{1}{n}$ does not converge:
  • Let $N$ be any positive integer. Choose integers $m, n$ such that $m > N$ and $n = 2m$.
  • Consider the absolute sum of terms from $m+1$ to $n$:       $|a_{m+1} + a_{m+2} + \dots + a_n| = \frac{1}{m+1} + \frac{1}{m+2} + \dots + \frac{1}{2m}$
  • Since each term satisfies $\frac{1}{m+k} > \frac{1}{2m}$ for $k < m$:       $\frac{1}{m+1} + \frac{1}{m+2} + \dots + \frac{1}{2m} > \frac{1}{2m} + \frac{1}{2m} + \dots + \frac{1}{2m} = m \left(\frac{1}{2m}\right) = \frac{1}{2}$
  • Choosing $\epsilon = \frac{1}{2}$, there exists no $N$ beyond which the block sum remains smaller than $\epsilon$, because for every $N$, integers $m, n$ ($n > m > N$) can be chosen such that:       $|a_{m+1} + a_{m+2} + \dots + a_n| > \frac{1}{2}$
  • By negation of Cauchy's general principle of convergence, the series $\sum_{n=1}^\infty \frac{1}{n}$ is not convergent.

Necessary Condition for Convergence and Boundedness

• Proposition (1.3) — Limit of Terms:

  • If the series $\sum a_n$ is convergent, then:     $\lim_{n \to \infty} (a_n) = 0$
  • Proof:
  • Let $S_n = a_1 + a_2 + \dots + a_n$.
  • Since $\sum a_n$ is convergent, the sequence ${S_n}$ converges to some limit $S$:       $\lim_{n \to \infty} (S_n) = S$
  • Express the $n^{\text{th}}$ term as the difference between consecutive partial sums:       $a_n = S_n - S_{n-1}$
  • Taking limits:       $\lim_{n \to \infty} (a_n) = \lim_{n \to \infty} (S_n) - \lim_{n \to \infty} (S_{n-1}) = S - S = 0$
  • Non-Sufficiency (Failure of the Converse):
  • The condition $\lim_{n \to \infty} (a_n) = 0$ is necessary but not sufficient for convergence.
  • Counterexample: For the series $\sum a_n = \sum \frac{1}{n}$:       $\lim_{n \to \infty} (a_n) = \lim_{n \to \infty} \left(\frac{1}{n}\right) = 0$       However, the series $\sum \frac{1}{n}$ is divergent.
  • Divergence Test (Contrapositive Note):
  • If $\lim_{n \to \infty} (a_n) \neq 0$, it definitively implies that the series $\sum a_n$ is not convergent.

• Theorem (1.2) — Series of Non-Negative Terms:

  • A series of non-negative terms ($a_n \ge 0$ for all $n$) converges if and only if its sequence of partial sums ${S_n}$ is bounded.
  • Proof:
  • Necessity: If $\sum a_n$ converges, the sequence ${S_n}$ is convergent and hence bounded.
  • Sufficiency: Let ${S_n}$ be bounded.
  • Since $a_n \ge 0$ for all $n$:       $S_{n+1} = S_n + a_{n+1} \ge S_n$
  • Thus, the sequence ${S_n}$ is monotonically increasing.
  • A monotonically increasing sequence that is bounded above converges.
  • Therefore, ${S_n}$ converges, meaning $\sum a_n$ is convergent.

Series of Positive Terms and Comparison Tests

• Dichotomy of Positive Term Series:

  • A series of positive terms $\sum a_n$ either converges or diverges to $+\infty$.
  • It cannot oscillate.

• Theorem (2.1) — Comparison Theorem:

  • Part (i): If $\sum a_n$ is a convergent series of positive terms and $\sum b_n$ is a series of positive terms such that $b_n \le k a_n$ for all $n$ (where $k > 0$), then $\sum b_n$ is also convergent.
  • Proof: Let $S_n = a_1 + a_2 + \dots + a_n$ and $S'_n = b_1 + b_2 + \dots + b_n$.
  • Then $b_1 + b_2 + \dots + b_n \le k(a_1 + a_2 + \dots + a_n)$, so $S'_n \le k S_n$.
  • Since $\sum a_n$ converges, ${S_n}$ is bounded. Therefore, ${S'_n}$ is bounded.
  • By Theorem (1.2), $\sum b_n$ is convergent.
  • Part (ii): If $\sum a_n$ is divergent and $b_n \ge k a_n$ for all $n$ (where $k > 0$), then $\sum b_n$ is also divergent.
  • Proof: If $\sum a_n$ is divergent, $\lim_{n \to \infty} (S_n) = +\infty$.
  • For every $k' > 0$, there exists $N$ such that $S_n > k'$ for all $n > N$.
  • Since $S'_n \ge k S_n$, it follows that $S'_n > k k'$ for $n > N$.
  • Hence $\lim_{n \to \infty} (S'_n) = +\infty$, meaning ${S'_n}$ diverges to $+\infty$, and $\sum b_n$ is divergent.

• Theorem (2.2) — Comparison Test (Limit Form):

  • Let $\sum a_n$ and $\sum b_n$ be two series of positive terms such that:     $\lim_{n \to \infty} \left(\frac{a_n}{b_n}\right) = l$
  • If $l$ is finite and non-zero ($l \neq 0$), then both series are either both convergent or both divergent.
  • If $l = 0$, the test fails.
  • Proof:
  • Since $\lim_{n \to \infty} \left(\frac{a_n}{b_n}\right) = l > 0$, choose $\epsilon$ such that $0 < \epsilon < l$.
  • There exists a positive integer $N$ such that for $n > N$:       $\left|\frac{a_n}{b_n} - l\right| < \epsilon \implies l - \epsilon < \frac{a_n}{b_n} < l + \epsilon$
  • Setting $k = l + \epsilon > 0$ and $k' = l - \epsilon > 0$:       $n > N \implies a_n < k b_n \quad \text{and} \quad a_n > k' b_n$
  • Removing the first $N$ terms from both series does not change their convergence nature.
  • By Comparison Theorem (2.1), if $\sum b_n$ converges, then $\sum a_n$ converges; if $\sum b_n$ diverges, then $\sum a_n$ diverges.
  • If $l = 0$, one cannot choose $\epsilon$ such that $0 < \epsilon < l$, so the argument fails.

The p-Series Test (Harmonic Series of Order p)

• Theorem (2.3) — Convergence Criteria for $\sum \frac{1}{n^p}$:

  • The series $\sum \frac{1}{n^p}$ converges if $p > 1$ and diverges if $p \le 1$.

• Proof by Grouping:

  • Case I ($p > 1$):
  • Because all terms are positive, terms can be grouped in brackets without altering the nature of the series:       $\frac{1}{1^p} + \left(\frac{1}{2^p} + \frac{1}{3^p}\right) + \left(\frac{1}{4^p} + \frac{1}{5^p} + \frac{1}{6^p} + \frac{1}{7^p}\right) + \left(\frac{1}{8^p} + \dots + \text{to } 8 \text{ terms}\right) + \dots \quad \text{--- (1)}$
  • Comparing this with the corresponding upper-bound groupings:       $\frac{1}{1^p} + \left(\frac{1}{2^p} + \frac{1}{2^p}\right) + \left(\frac{1}{4^p} + \frac{1}{4^p} + \frac{1}{4^p} + \frac{1}{4^p}\right) + \left(\frac{1}{8^p} + \dots + \text{to } 8 \text{ terms}\right) + \dots$$= \frac{1}{1^p} + 2\left(\frac{1}{2^p}\right) + 4\left(\frac{1}{4^p}\right) + 8\left(\frac{1}{8^p}\right) + \dots$$= \frac{1}{1^p} + \frac{1}{2^{p-1}} + \left(\frac{1}{2^{p-1}}\right)^2 + \left(\frac{1}{2^{p-1}}\right)^3 + \dots \quad \text{--- (2)}$
  • The terms of grouped series (1) are less than the corresponding terms of series (2) after the first term.
  • Series (2) is a geometric series with common ratio $r = \frac{1}{2^{p-1}}$.
  • Since $p > 1$, $p - 1 > 0$, so $2^{p-1} > 1$, which implies $r = \frac{1}{2^{p-1}} < 1$.
  • Hence, series (2) converges. By Comparison Theorem (2.1) (with $k = 1$), series (1) converges, and $\sum \frac{1}{n^p}$ is convergent.
  • Case II ($p = 1$):
  • The series becomes the harmonic series:       $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots$
  • Grouping terms:       $1 + \frac{1}{2} + \left(\frac{1}{3} + \frac{1}{4}\right) + \left(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}\right) + \dots \quad \text{--- (3)}$
  • Comparing with:       $1 + \frac{1}{2} + \left(\frac{1}{4} + \frac{1}{4}\right) + \left(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}\right) + \dots$$= 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \dots \quad \text{--- (4)}$
  • The terms of series (3) are strictly greater than the corresponding terms of series (4) after the second term.
  • Series (4) is divergent because $\lim_{n \to \infty} (a_n) = \frac{1}{2} \neq 0$.
  • By Comparison Theorem (2.1), grouped series (3) diverges, so $\sum \frac{1}{n}$ is divergent.
  • Case III ($p < 1$):
  • For all $n > 1$, $n^p < n$, which implies:       $\frac{1}{n^p} > \frac{1}{n}$
  • The terms of $\sum \frac{1}{n^p}$ are greater than the corresponding terms of $\sum \frac{1}{n}$ after the first term.
  • Since $\sum \frac{1}{n}$ diverges (by Case II), by Comparison Theorem (2.1), $\sum \frac{1}{n^p}$ is divergent when $p < 1$.

Worked Examples for Comparison Testing

• Example-Set (2.1) (i):

  • Series: $\sum a_n$ where $a_n = \sqrt{n^4 + 1} - \sqrt{n^4 - 1}$
  • Factor out $n^2$:     $a_n = n^2 \left(1 + \frac{1}{n^4}\right)^{1/2} - n^2 \left(1 - \frac{1}{n^4}\right)^{1/2}$
  • Using the binomial expansion:     $a_n = n^2 \left[1 + \frac{1}{2n^4} - \frac{1}{8n^8} + \dots\right] - n^2 \left[1 - \frac{1}{2n^4} - \frac{1}{8n^8} - \dots\right]$$a_n = n^2 \left[\frac{1}{n^4} + \frac{B}{n^{12}} + \dots\right] = \frac{1}{n^2} \left[1 + \frac{B}{n^8} + \dots\right]$
  • Choose $b_n = \frac{1}{n^2}$. Taking the limit:     $\lim_{n \to \infty} \left(\frac{a_n}{b_n}\right) = 1$
  • The series $\sum b_n = \sum \frac{1}{n^2}$ is convergent by Theorem (2.3) ($p = 2 > 1$).
  • Hence, by Comparison Test (2.2), $\sum a_n$ is convergent.

• Example-Set (2.1) (ii):

  • Series: $\sum a_n$ where $a_n = \sqrt{n^2 + 1} - n$
  • Factor out $n$:     $a_n = n \left[\left(1 + \frac{1}{n^2}\right)^{1/2} - 1\right]$
  • Using the binomial expansion:     $a_n = n \left[1 + \frac{1}{2n^2} - \frac{1}{8n^4} + \dots - 1\right] = \frac{1}{n} \left[\frac{1}{2} - \frac{1}{8n^2} + \dots\right]$
  • Choose $b_n = \frac{1}{n}$. Taking the limit:     $\lim_{n \to \infty} \left(\frac{a_n}{b_n}\right) = \frac{1}{2}$
  • The series $\sum b_n = \sum \frac{1}{n}$ is divergent by Theorem (2.3) ($p = 1$).
  • Therefore, by Comparison Test (2.2), $\sum a_n$ is divergent.

D'Alembert's Ratio Test (First Ratio Test)

• Theorem (2.4) — Standard Ratio Formulation:

  • Let $\sum a_n$ be a series of positive terms such that:     $\lim_{n \to \infty} \left(\frac{a_{n+1}}{a_n}\right) = l$
  • (i) $\sum a_n$ is convergent if $l < 1$.
  • (ii) $\sum a_n$ is divergent if $l > 1$.
  • If $l = 1$, the test fails.

• Proof:

  • Case (i) ($l < 1$):
  • Choose $\epsilon > 0$ such that $r = l + \epsilon < 1$.
  • Since $\lim_{n \to \infty} \left(\frac{a_{n+1}}{a_n}\right) = l$, there exists a positive integer $N$ such that for all $n > N$:       $\frac{a_{n+1}}{a_n} < l + \epsilon = r$
  • Setting $n = N+1, N+2, N+3, \dots$:       $a_{N+2} < r a_{N+1}$$a_{N+3} < r a_{N+2} < r^2 a_{N+1}$$a_{N+p} < r^{p-1} a_{N+1} \quad \text{where } p > 1$
  • The terms of the tail series $\sum_{p=2}^\infty a_{N+p}$ are less than $k (= a_{N+1})$ times the corresponding terms of the geometric series $\sum_{p=1}^\infty r^p$.
  • Since $r < 1$, the geometric series converges.
  • By Comparison Theorem (2.1), $\sum_{p=2}^\infty a_{N+p}$ converges, and consequently $\sum_{n=1}^\infty a_n$ is convergent.
  • Case (ii) ($l > 1$):
  • Choose $\epsilon_1 > 0$ such that $R = l - \epsilon_1 > 1$.
  • There exists a positive integer $N$ such that for all $n > N$:       $R = l - \epsilon_1 < \frac{a_{n+1}}{a_n}$
  • Successive substitution yields:       $a_{N+2} > R a_{N+1}$$a_{N+3} > R a_{N+2} > R^2 a_{N+1}$$a_{N+p} > R^{p-1} a_{N+1} \quad \text{when } p > 1$
  • The terms of $\sum_{p=2}^\infty a_{N+p}$ are greater than $k (= a_{N+1})$ times the terms of $\sum_{p=1}^\infty R^p$.
  • Since $R > 1$, the geometric series diverges.
  • By Comparison Theorem (2.1), $\sum_{p=2}^\infty a_{N+p}$ diverges, and $\sum_{n=1}^\infty a_n$ is divergent.
  • Case ($l = 1$):
  • One cannot choose $\epsilon > 0$ such that $1 + \epsilon < 1$ or $1 - \epsilon > 1$, causing the argument to fail.

• Note — Alternative Form (First Ratio Test):

  • If $\sum a_n$ is a series of positive terms and:     $\lim_{n \to \infty} \left(\frac{a_n}{a_{n+1}}\right) = L$
  • (i) $\sum a_n$ is convergent if $L > 1$.
  • (ii) $\sum a_n$ is divergent if $L < 1$.
  • If $L = 1$, the test fails.

Comparison of Ratios Theorem

• Theorem (2.5) Statement:

  • Let $\sum a_n$ and $\sum b_n$ be two series of positive terms such that:     $\frac{a_n}{a_{n+1}} > \frac{b_n}{b_{n+1}} \quad \text{for all } n > m \text{ (where } m \text{ is a fixed positive integer)}$
  • (i) $\sum a_n$ is convergent if $\sum b_n$ converges.
  • (ii) $\sum b_n$ is divergent if $\sum a_n$ diverges.

• Proof:

  • Let partial tail sums be defined as:     $S_p = a_{m+1} + a_{m+2} + \dots + a_{m+p}$$S'_p = b_{m+1} + b_{m+2} + \dots + b_{m+p}$
  • Factoring out initial terms:     $S_p = a_{m+1} \left(1 + \frac{a_{m+2}}{a_{m+1}} + \frac{a_{m+3}}{a_{m+2}} \cdot \frac{a_{m+2}}{a_{m+1}} + \dots\right)$
  • Because $\frac{a_{n+1}}{a_n} < \frac{b_{n+1}}{b_n}$ for all indices greater than $m$:     $S_p < a_{m+1} \left(1 + \frac{b_{m+2}}{b_{m+1}} + \frac{b_{m+3}}{b_{m+2}} \cdot \frac{b_{m+2}}{b_{m+1}} + \dots\right)$$S_p < \frac{a_{m+1}}{b_{m+1}} (b_{m+1} + b_{m+2} + b_{m+3} + \dots + b_{m+p}) = \frac{a_{m+1}}{b_{m+1}} S'_p$
  • Thus $S_p < k S'_p$, where $k = \frac{a_{m+1}}{b_{m+1}}$.
  • If $\sum_{p=1}^\infty b_{m+p}$ is convergent, the sequence ${S'_p}$ is bounded.
  • Consequently, ${S_p}$ is bounded, establishing that $\sum_{p=1}^\infty a_{m+p}$ and hence $\sum a_n$ converges.
  • Part (ii) follows analogously by contraposition.

Raabe's Test

• Theorem (2.6) Statement:

  • Let $\sum a_n$ be a series of positive terms and:     $\lim_{n \to \infty} \left(n \left(\frac{a_n}{a_{n+1}} - 1\right)\right) = l$
  • (i) $\sum a_n$ is convergent if $l > 1$.
  • (ii) $\sum a_n$ is divergent if $l < 1$.
  • If $l = 1$, the test fails.

• Proof:

  • Part (i) ($l > 1$):
  • Choose $p$ such that $1 < p < l$.
  • Select the comparison series $b_n = \frac{1}{n^p}$, which converges by Theorem (2.3) since $p > 1$.
  • By Theorem (2.5) (Comparison of Ratios), $\sum a_n$ converges if:       $\frac{a_n}{a_{n+1}} > \frac{b_n}{b_{n+1}} = \frac{(n+1)^p}{n^p} = \left(1 + \frac{1}{n}\right)^p$
  • Expanding via the binomial theorem:       $\frac{a_n}{a_{n+1}} > 1 + \frac{p}{n} + \frac{p(p-1)}{2! n^2} + \dots$$n \left(\frac{a_n}{a_{n+1}} - 1\right) > p + \frac{A}{n} + \frac{B}{n^2} + \dots$
  • Taking the limit as $n \to \infty$:       $\lim_{n \to \infty} \left(n \left(\frac{a_n}{a_{n+1}} - 1\right)\right) > p + \lim_{n \to \infty} \left(\frac{a}{n} + \frac{b}{n^2} + \dots\right)$$\lim_{n \to \infty} \left(n \left(\frac{a_n}{a_{n+1}} - 1\right)\right) > p$
  • This condition holds because $l > p > 1$. Therefore, $\sum a_n$ is convergent.
  • Part (ii) ($l < 1$):
  • Analogously, comparing with $b_n = \frac{1}{n^p}$ with $p < 1$ establishes divergence.
  • Failure at $l = 1$:
  • If $l = 1$, one cannot choose $p$ strictly between $1$ and $l$, so the test fails.

• Operational Context:

  • Raabe's test is applied when D'Alembert's Ratio Test fails (i.e., when $\lim_{n \to \infty} \left(\frac{a_n}{a_{n+1}}\right) = 1$).

Logarithmic Ratio Test

• Theorem (2.7) Statement:

  • A series $\sum a_n$ of positive terms is:
  • (i) Convergent if:       $\lim_{n \to \infty} \left(n \log\left(\frac{a_n}{a_{n+1}}\right)\right) > 1$
  • (ii) Divergent if:       $\lim_{n \to \infty} \left(n \log\left(\frac{a_n}{a_{n+1}}\right)\right) < 1$
  • The test fails if the limit is $1$.

• Proof:

  • Let $b_n = \frac{1}{n^p}$ with $p > 1$, so $\sum b_n$ is convergent.
  • By Comparison of Ratios Theorem (2.5), $\sum a_n$ is convergent if for every $n > m$:     $\frac{a_n}{a_{n+1}} > \frac{b_n}{b_{n+1}} = \left(1 + \frac{1}{n}\right)^p$
  • Taking the natural logarithm of both sides:     $\log\left(\frac{a_n}{a_{n+1}}\right) > p \log\left(1 + \frac{1}{n}\right) = p \left(\frac{1}{n} - \frac{1}{2n^2} + \frac{1}{3n^3} - \dots\right)$$\log\left(\frac{a_n}{a_{n+1}}\right) > \frac{p}{n} - \frac{p}{2n^2} + \frac{p}{3n^3} - \dots$
  • Multiplying across by $n$:     $n \log\left(\frac{a_n}{a_{n+1}}\right) > p - \frac{p}{2n} + \frac{p}{3n^2} - \dots$
  • Taking limits:     $\lim_{n \to \infty} \left(n \log\left(\frac{a_n}{a_{n+1}}\right)\right) > p - \lim_{n \to \infty} \left(\frac{p}{2n} - \frac{p}{3n^2} + \dots\right) = p$
  • Since $p > 1$, convergence is guaranteed if $\lim_{n \to \infty} \left(n \log\left(\frac{a_n}{a_{n+1}}\right)\right) > 1$.
  • The divergence condition follows similarly.

• Application Note:

  • The Logarithmic Ratio Test serves as an alternative to Raabe's Test.
  • It is particularly suitable when $\lim_{n \to \infty} \left(\frac{a_n}{a_{n+1}}\right)$ evaluates to an expression involving the base $e$, or when evaluating $\lim_{n \to \infty} \left(n \log\left(\frac{a_n}{a_{n+1}}\right)\right)$ is straightforward.

De Morgan's and Bertrand's Test

• Theorem (2.8) Statement:

  • If $\sum a_n$ is a series of positive terms, the series is convergent or divergent according as:     $\lim_{n \to \infty} \left(\left\{n \left(\frac{a_n}{a_{n+1}} - 1\right) - 1\right\} \log(n)\right) > 1 \quad \text{or} \quad < 1$

• Proof:

  • Use the comparison series $b_n = \frac{1}{n (\log(n))^p}$, which converges if $p > 1$ and diverges if $p \le 1$.
  • Compute the ratio of consecutive terms:     $\frac{b_n}{b_{n+1}} = \frac{n+1}{n} \left\{\frac{\log(n+1)}{\log(n)}\right\}^p = \left(1 + \frac{1}{n}\right) \left\{\frac{\log(n) + \log\left(1 + \frac{1}{n}\right)}{\log(n)}\right\}^p$$= \left(1 + \frac{1}{n}\right) \left\{1 + \frac{1}{\log(n)} \left(\frac{1}{n} - \frac{1}{2n^2} + \dots\right)\right\}^p$$= \left(1 + \frac{1}{n}\right) \left\{1 + \frac{p}{n \log(n)} + \dots\right\} = 1 + \frac{1}{n} + \frac{p}{n \log(n)} + \dots$
  • By Comparison of Ratios Theorem, $\sum a_n$ converges if for $n > m$ and $p > 1$:     $\frac{a_n}{a_{n+1}} > 1 + \frac{1}{n} + \frac{p}{n \log(n)} + \dots$$n \left(\frac{a_n}{a_{n+1}} - 1\right) > 1 + \frac{p}{\log(n)} + \dots$$\left\{n \left(\frac{a_n}{a_{n+1}} - 1\right) - 1\right\} \log(n) > p + \text{terms containing } n \text{ and } \log(n) \text{ in denominator}$
  • Taking the limit as $n \to \infty$:     $\lim_{n \to \infty} \left(\left\{n \left(\frac{a_n}{a_{n+1}} - 1\right) - 1\right\} \log(n)\right) > p > 1$
  • Thus, $\sum a_n$ is convergent if the limit is strictly greater than $1$.

Cauchy's Root Test

• Theorem (2.9) Statement:

  • If $\sum a_n$ is a series of positive terms such that:     $\lim_{n \to \infty} (a_n)^{1/n} = l$
  • $\sum a_n$ is convergent if $l < 1$.
  • $\sum a_n$ is divergent if $l > 1$.
  • The test fails if $l = 1$.

• Proof:

  • Case (i) ($l < 1$):
  • Choose $\epsilon > 0$ such that $l + \epsilon < 1$.
  • Since $\lim_{n \to \infty} (a_n)^{1/n} = l$, there exists a positive integer $N$ such that for $n > N$:       $(a_n)^{1/n} < l + \epsilon = r \quad \text{where } r < 1$$a_n < r^n$
  • The terms of $\sum_{n=N+1}^\infty a_n$ are less than the corresponding terms of the geometric series $\sum_{n=N+1}^\infty r^n$.
  • Since $r < 1$, the geometric series converges.
  • By Comparison Theorem, $\sum a_n$ is convergent.
  • Case (ii) ($l > 1$):
  • Choose $\epsilon > 0$ such that $l - \epsilon > 1$.
  • There exists a positive integer $N$ such that for $n > N$:       $(a_n)^{1/n} > l - \epsilon = R \quad \text{where } R > 1$$a_n > R^n$
  • Since $\sum R^n$ diverges for $R > 1$, by Comparison Theorem, $\sum a_n$ is divergent.
  • Case (iii) ($l = 1$):
  • One cannot choose $\epsilon > 0$ such that $l + \epsilon < 1$ or $l - \epsilon > 1$, so the argument fails.

Cauchy's Condensation Test

• Theorem (2.10) Statement:

  • Let $\phi$ be a function such that:
  • (i) $\phi(n)$ is positive for all positive integers $n$.
  • (ii) The sequence ${\phi(n)}$ is non-increasing.
  • Then the two series:     $\sum \phi(n) \quad \text{and} \quad \sum a^n \phi(a^n)$     are either both convergent or both divergent, where $a$ is a positive integer greater than $1$.

• Proof:

  • Consider the block sum:     $\phi(a^k + 1) + \phi(a^k + 2) + \dots + \phi(a^{k+1})$
  • This group consists of $a^{k+1} - a^k = a^k (a - 1)$ terms.
  • Because $\phi(n)$ is non-increasing, each term is greater than or equal to $\phi(a^{k+1})$ and less than or equal to $\phi(a^k)$.
  • Therefore:     $(a - 1) a^k \phi(a^{k+1}) \le \phi(a^k + 1) + \phi(a^k + 2) + \dots + \phi(a^{k+1}) \le (a - 1) a^k \phi(a^k)$
  • Evaluating this inequality successively for $k = 1, 2, \dots, p$:     $(a - 1) a \phi(a^2) \le \phi(a + 1) + \dots + \phi(a^2) \le (a - 1) a \phi(a)$$(a - 1) a^2 \phi(a^3) \le \phi(a^2 + 1) + \dots + \phi(a^3) \le (a - 1) a^2 \phi(a^2)$$\dots$$(a - 1) a^p \phi(a^{p+1}) \le \phi(a^p + 1) + \dots + \phi(a^{p+1}) \le (a - 1) a^p \phi(a^p)$
  • Summing these inequalities yields inequality (1):     $\frac{a - 1}{a} \left[a^2 \phi(a^2) + a^3 \phi(a^3) + \dots + a^{p+1} \phi(a^{p+1})\right] \le \phi(a + 1) + \phi(a + 2) + \dots + \phi(a^{p+1})$$\le (a - 1) \left[a \phi(a) + a^2 \phi(a^2) + \dots + a^p \phi(a^p)\right]$
  • Convergence of $\sum a^n \phi(a^n)$ when $\sum \phi(n)$ converges:
  • Let $\sum \phi(n)$ converge. Its partial sums are bounded: there exists $A$ such that:       $\phi(1) + \phi(2) + \dots + \phi(n) < A \quad \text{for all } n$
  • By the left side of inequality (1):       $\frac{a - 1}{a} \left[a^2 \phi(a^2) + \dots + a^{p+1} \phi(a^{p+1})\right] < A$$a^2 \phi(a^2) + \dots + a^{p+1} \phi(a^{p+1}) < \frac{a}{a - 1} A + a \phi(a)$
  • Thus, the partial sums of $\sum a^n \phi(a^n)$ are bounded for all $p$, meaning $\sum a^n \phi(a^n)$ converges.
  • Convergence of $\sum \phi(n)$ when $\sum a^n \phi(a^n)$ converges:
  • Let $\sum a^n \phi(a^n)$ converge. Its partial sums are bounded: there exists $B$ such that:       $a \phi(a) + a^2 \phi(a^2) + \dots + a^n \phi(a^n) < B \quad \text{for all } n$
  • By the right side of inequality (1):       $\phi(a + 1) + \phi(a + 2) + \dots + \phi(a^{p+1}) < (a - 1) B$$\phi(1) + \phi(2) + \dots + \phi(a^{p+1}) < (a - 1) B + \phi(1) + \dots + \phi(a)$
  • For any positive integer $n$, choose $p$ such that $n < a^{p+1}$. Then:       $\phi(1) + \phi(2) + \dots + \phi(n) < \phi(1) + \phi(2) + \dots + \phi(a^{p+1}) < (a - 1) B + \phi(1) + \dots + \phi(a)$
  • The sequence of partial sums of $\sum \phi(n)$ is bounded, so $\sum \phi(n)$ converges.