Comprehensive Notes on Infinite Series and Convergence Tests

Definition of Infinite Series

• Definition 1 (Infinite Series): An infinite (numerical) series, or simply a series, is an expression that can be written in the form:     $\sum_{k=1}^{+\infty} u_k = u_1 + u_2 + u_3 + \dots + u_k + \dots$

• The numbers $u_1, u_2, u_3, \dots$ are called the terms of the series.

• Since it is impossible to add an infinite number of terms directly, the sum of an infinite series is determined using an indirect limit process.

• Conceptual Example: Consider the decimal number $0.3333333\dots$

  • This number can be written as the infinite series: $0.3 + 0.03 + 0.003 + 0.0003 + \dots$

  • Equivalently, this is the sum: $\frac{3}{10} + \frac{3}{10^2} + \frac{3}{10^3} + \frac{3}{10^4} + \dots$

  • Since $0.3333333\dots$ is the decimal expansion of $\frac{1}{3}$, any reasonable definition for the sum of an infinite series should yield the result $\frac{1}{3}$ for this sum.

Sequence of Partial Sums and Convergence

• Definitions of Partial Sums: To formalize the sum, let $s_n$ denote the sum of the first $n$ terms of the series:

  • $s_1 = u_1$

  • $s_2 = u_1 + u_2$

  • $s_3 = u_1 + u_2 + u_3$

  • $s_n = \sum_{k=1}^{n} u_k = u_1 + u_2 + u_3 + \dots + u_n$

• The number $s_n$ is called the $n$-th partial sum of the series.

• The sequence $\{s_n\}_{n=1}^{+\infty}$ is called the sequence of partial sums of the series.

• Definition 2 (Convergence and Divergence): Let $\{s_n\}$ be the sequence of partial sums of the series $\sum_{k=1}^{+\infty} u_k$.

  • If the sequence $\{s_n\}$ converges to a real number $S$, then we say that the series converges to $S$. $S$ is called the sum of the series, denoted as $\sum_{k=1}^{+\infty} u_k = S$.

  • If the sequence of partial sums of the series diverges, we say that the series diverges. A divergent series does not have a sum.

Example 1: Oscillating Series

• Determine whether the series $1 - 1 + 1 - 1 + 1 - 1 + \dots$ converges or diverges.

• Solution: Calculate the partial sums:

  • $s_1 = 1$

  • $s_2 = 1 - 1 = 0$

  • $s_3 = 1 - 1 + 1 = 1$

  • $s_4 = 1 - 1 + 1 - 1 = 0$

• The sequence of partial sums is $1, 0, 1, 0, 1, 0, \dots$.

• Since this sequence is divergent (it oscillates), the given series diverges and has no sum.

Geometric Series

• Definition: A geometric series is a series in which each term is obtained by multiplying the previous term by a fixed constant $r$. If the starting term is $a \neq 0$, the series has the form:     $\sum_{k=0}^{+\infty} ar^k = a + ar + ar^2 + ar^3 + \dots + ar^k + \dots$

• The constant $r$ is called the ratio or quotient of the series.

• Theorem 1 (Convergence of Geometric Series): A geometric series converges if |r| < 1 and diverges if $|r| \geq 1$. If the series converges, its sum is given by:     $\sum_{k=0}^{+\infty} ar^k = \frac{a}{1 - r}$

Proof of Theorem 1 (Geometric Series)

• Case 1 ($|r| = 1$):

  • If $r = 1$, the series is $a + a + a + a + \dots$. The $n$-th partial sum is $s_n = (n + 1)a$.

  • $\lim_{n \to +\infty} s_n = \lim_{n \to +\infty} (n + 1)a$. This limit is $+\infty$ if a > 0 and $-\infty$ if a < 0. In both cases, the series diverges.

  • If $r = -1$, the series is $a - a + a - a + \dots$. The sequence of partial sums is $a, 0, a, 0, \dots$, which diverges. Thus, the series diverges.

• Case 2 ($|r| \neq 1$):

  • The $n$-th partial sum is $s_n = a + ar + ar^2 + \dots + ar^n$.

  • Multiply by $r$: $rs_n = ar + ar^2 + \dots + ar^n + ar^{n+1}$.

  • Subtract: $s_n - rs_n = a - ar^{n+1} \implies (1 - r)s_n = a(1 - r^{n+1})$.

  • Since $r \neq 1$, we have $s_n = \frac{a(1 - r^{n+1})}{1 - r}$.

  • If |r| < 1, then $r^{n+1} \to 0$ as $n \to +\infty$. Thus, $\lim_{n \to +\infty} s_n = \frac{a}{1 - r}$.

  • If |r| > 1, then if r > 1, $r^{n+1}$ grows unbounded. If r < -1, $r^{n+1}$ oscillates between increasing positive and negative values. In both cases, the sequence $\{s_n\}$ diverges.

Example 2: Testing Geometric Series

• (a) Series: $\sum_{k=0}^{+\infty} 5\left(\frac{1}{4}\right)^k$

  • Solution: This is a geometric series with $a = 5$ and $r = \frac{1}{4}$.

  • Since |r| = \frac{1}{4} < 1, the series converges.

  • Sum: $S = \frac{5}{1 - \frac{1}{4}} = \frac{5}{\frac{3}{4}} = \frac{20}{3}$.

• (b) Series: $\sum_{k=1}^{+\infty} \frac{3^{2k-1}}{5^{k+1}}$

  • Solution: Rewrite the general term: $\frac{3^{2k} \cdot 3^{-1}}{5^k \cdot 5} = \frac{1}{15} \cdot \frac{9^k}{5^k} = \frac{1}{15} \left(\frac{9}{5}\right)^k$.

  • This is a geometric series with $r = \frac{9}{5}$.

  • Since r = \frac{9}{5} > 1, the series diverges.

Telescoping Series

• Definition: A series of the form $\sum_{k=1}^{+\infty} \frac{1}{k(k+1)}$ is known as a telescoping series.

• Evaluation: To check convergence, rewrite the general term using partial fractions:     $\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$

• The $n$-th partial sum $s_n$ is:     $s_n = \sum_{k=1}^{n} \left(\frac{1}{k} - \frac{1}{k+1}\right) = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \dots + \left(\frac{1}{n} - \frac{1}{n+1}\right)$

• After cancellation (telescoping): $s_n = 1 - \frac{1}{n + 1}$.

• The limit is: $\lim_{n \to +\infty} s_n = \lim_{n \to +\infty} \left(1 - \frac{1}{n+1}\right) = 1$.

• The series converges to $1$.

Harmonic Series

• Definition: The series of the form $\sum_{k=1}^{+\infty} \frac{1}{k} = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{k} + \dots$ is called the harmonic series.

• It is one of the most important divergent series.

• Proof of Divergence:

  • The terms are positive, so the sequence of partial sums is strictly increasing: s_1 < s_2 < s_3 < \dots < s_n < \dots

  • By examining selected partial sums (where the index is a power of 2):

    • $s_2 = 1 + \frac{1}{2}$

    • s_4 = 1 + \frac{1}{2} + (\frac{1}{3} + \frac{1}{4}) > 1 + \frac{1}{2} + (\frac{1}{4} + \frac{1}{4}) = 1 + \frac{2}{2}

    • s_8 = s_4 + (\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}) > (1 + \frac{2}{2}) + (\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}) = 1 + \frac{3}{2}

  • In general, s_{2^n} > 1 + \frac{n}{2}.

  • As $n \to +\infty$, the value $1 + \frac{n}{2}$ grows without bound. Therefore, the sequence of partial sums is not bounded from above.

  • Consequently, the harmonic series diverges.

Divergence Test

• Theorem 2: If the series $\sum u_k$ converges, then $\lim_{k \to +\infty} u_k = 0$.

  • Proof: Since $s_k = s_{k-1} + u_k$, we have $u_k = s_k - s_{k-1}$.

  • If the series converges and $\lim_{k \to +\infty} s_k = S$, then $\lim_{k \to +\infty} s_{k-1} = S$ as well.

  • Therefore, $\lim_{k \to +\infty} u_k = \lim_{k \to +\infty} (s_k - s_{k-1}) = S - S = 0$.

• Theorem 3 (Divergence Test):

  • (i) If $\lim_{k \to +\infty} u_k \neq 0$, then the series $\sum u_k$ diverges.

  • (ii) If $\lim_{k \to +\infty} u_k = 0$, the series may converge or diverge (the test is inconclusive).

  • Example 3: Examine the convergence of $\sum_{k=1}^{+\infty} \frac{k}{k+1}$.

    • $\lim_{k \to +\infty} \frac{k}{k+1} = 1 \neq 0$. Therefore, the series diverges.

Algebraic Properties of Infinite Series

Theorem 4:

• (i) If $\sum u_k$ and $\sum v_k$ are convergent series, then $\sum (u_k + v_k)$ and $\sum (u_k - v_k)$ are convergent series, and their sums are:

  • $\sum (u_k + v_k) = \sum u_k + \sum v_k$

  • $\sum (u_k - v_k) = \sum u_k - \sum v_k$

• (ii) If $c \neq 0$, the series $\sum cu_k$ and $\sum u_k$ converge or diverge simultaneously, and $\sum cu_k = c \sum u_k$.

• (iii) Deleting or adding a finite number of terms does not affect convergence or divergence of the series. For any constant $K$, the series $\sum_{k=1}^{+\infty} u_k$ and $\sum_{k=K}^{+\infty} u_k$ share the same nature.

Example 4: Find the sum of $\sum_{k=1}^{+\infty} \left( \frac{3}{4^k} - \frac{2}{5^{k-1}} \right)$.

• Series 1 ($a = \frac{3}{4}, r = \frac{1}{4}$): Sum $= \frac{3/4}{1 - 1/4} = 1$.

• Series 2 ($a = 2, r = \frac{1}{5}$): Sum $= \frac{2}{1 - 1/5} = \frac{2}{4/5} = \frac{10}{4} = 2.5$.

• Result: $1 - 2.5 = -1.5$.

Integral Test

Theorem 5 (Integral Test):

• Let $\sum u_k$ be a series with positive terms.

• If $f$ is a positive, decreasing, and continuous function on the interval $[a, +\infty)$, where $a \in \mathbb{Z}$ and $u_k = f(k)$ for every $k \geq a$, then:

  • $\sum_{k=a}^{+\infty} u_k$ and $\int_{a}^{+\infty} f(x)dx$ have the same nature (both converge or both diverge).

Example 6 (Applying Integral Test):

• (a) Harmonic Series $\sum_{k=1}^{+\infty} \frac{1}{k}$: Let $f(x) = \frac{1}{x}$. This is positive, decreasing, and continuous for $x \geq 1$.

  • $\int_{1}^{+\infty} \frac{1}{x}dx = [\ln|x|]_{1}^{+\infty} = \lim_{l \to +\infty} \ln(l) - 0 = +\infty$. The series diverges.

• (b) Series $\sum_{k=1}^{+\infty} \frac{1}{k^2}$: Let $f(x) = \frac{1}{x^2}$. This is positive, decreasing, and continuous for $x \geq 1$.

  • $\int_{1}^{+\infty} \frac{1}{x^2}dx = [-\frac{1}{x}]_{1}^{+\infty} = 0 - (-1) = 1$. The series converges (though this integral value is not the sum of the series).

p-series (Hyperharmonic Series)

• Definition: A series of the form:     $\sum_{k=1}^{+\infty} \frac{1}{k^p} = 1 + \frac{1}{2^p} + \frac{1}{3^p} + \dots + \frac{1}{k^p} + \dots$     where p > 0 is called a p-series or hyperharmonic series.

• Theorem 6 (Convergence of a p-series):

  • A p-series converges if p > 1.

  • A p-series diverges if 0 < p \leq 1.

• Proof (using Integral Test for $p \neq 1$):

  • $\int_{1}^{+\infty} \frac{1}{x^p}dx = \lim_{l \to +\infty} [\frac{x^{1-p}}{1-p}]_{1}^{l} = \lim_{l \to +\infty} \left( \frac{l^{1-p}}{1-p} - \frac{1}{1-p} \right)$.

  • If p > 1, then 1 - p < 0, so $l^{1-p} \to 0$. The integral converges to $\frac{-1}{1-p}$.

  • If 0 < p < 1, then 1 - p > 0, so $l^{1-p} \to +\infty$. The integral diverges.

  • For $p = 1$, it is the harmonic series, which diverges.

• Example 7: The series $\sum_{k=1}^{+\infty} \frac{1}{\sqrt[3]{k}} = \sum_{k=1}^{+\infty} \frac{1}{k^{1/3}}$ diverges because it is a p-series with $p = \frac{1}{3} \leq 1$.