Infinite Sequences and Series - Convergence Tests

Geometric Series

• A geometric series can be written in the form $a ^{n}$, where:

  • $r$ is the common ratio.

  • $a$ is the first term.

• A geometric series converges if \left| r \right| < 1.

• If a geometric series converges, its sum is given by: $\frac{a}{1 - r}$

• To find $a$, always plug in the initial index value into the expression.

Examples of Geometric Series

1. $\sum_{n=0}^{\infty} 3 \left( \frac{1}{2} \right)^{n}$

   • $r = \frac{1}{2}$

   • Since \left| \frac{1}{2} \right| < 1, the series converges.

   • $a = 3 \left( \frac{1}{2} \right)^{0} = 3$

   • Sum: $\frac{3}{1 - \frac{1}{2}} = \frac{3}{\frac{1}{2}} = 6$

2. $\sum_{n=1}^{\infty} \frac{3}{4^{n}}$

   • Rewrite as: $3 \sum_{n=1}^{\infty} \left( \frac{1}{4} \right)^{n}$

   • $r = \frac{1}{4}$

   • Since \left| \frac{1}{4} \right| < 1, the series converges.

   • $a = 3 \left( \frac{1}{4} \right)^{1} = \frac{3}{4}$

   • Sum: $\frac{\frac{3}{4}}{1 - \frac{1}{4}} = \frac{\frac{3}{4}}{\frac{3}{4}} = 1$

3. $\sum_{n=1}^{\infty} \left( \frac{\pi}{3} \right)^{n}$

   • r = \frac{\pi}{3} > 1

   • Since \left| \frac{\pi}{3} \right| > 1, the series diverges.

4. $\sum_{n=1}^{\infty} \frac{2}{(-3)^{n}}$

   • Rewrite as: $2 \sum_{n=1}^{\infty} \left( -\frac{1}{3} \right)^{n}$

   • $r = -\frac{1}{3}$

   • Since \left| -\frac{1}{3} \right| < 1, the series converges.

   • $a = 2 \left( -\frac{1}{3} \right)^{1} = -\frac{2}{3}$

   • Sum: $\frac{-\frac{2}{3}}{1 - \left( -\frac{1}{3} \right)} = \frac{-\frac{2}{3}}{\frac{4}{3}} = -\frac{1}{2}$

nth Term Test for Divergence

• If $\lim<em>{n \to \infty} a</em>{n} \neq 0$, then the series diverges.

• If $\lim<em>{n \to \infty} a</em>{n} = 0$, the test is inconclusive.

• Example:

  • $\sum_{n=2}^{\infty} \frac{2n + 1}{3n - 1}$

  • $\lim_{n \to \infty} \frac{2n + 1}{3n - 1} = \frac{2}{3} \neq 0$

  • Therefore, the series diverges by the nth term test.

Integral Test and P-Series

Integral Test

• Conditions:

  • $f(x)$ is continuous on $\left[ k, \infty \right)$, for some integer $k$.

  • $f(x)$ is positive.

  • $f(x)$ is decreasing.

• If these conditions hold, then $\sum<em>{n=k}^{\infty} a</em>{n}$ and $\int_{k}^{\infty} f(x) dx$ either both converge or both diverge.

• Example 1:

  • $\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{2}}$

  • $f(x) = \frac{1}{x(\ln x)^{2}}$

  • $\lim<em>{R \to \infty} \int</em>{2}^{R} \frac{1}{x(\ln x)^{2}} dx$

    • Let $u = \ln x$, $du = \frac{1}{x} dx$

    • $\int \frac{1}{u^{2}} du = -\frac{1}{u} = -\frac{1}{\ln x}$

  • $\lim<em>{R \to \infty} \left[ -\frac{1}{\ln x} \right]</em>{2}^{R} = \lim_{R \to \infty} \left( -\frac{1}{\ln R} + \frac{1}{\ln 2} \right) = \frac{1}{\ln 2}$

  • Since the integral converges, the series converges.

• Example 2:

  • $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n + 4}}$

  • $f(x) = (x + 4)^{-\frac{1}{2}}$

  • $\lim<em>{R \to \infty} \int</em>{1}^{R} (x + 4)^{-\frac{1}{2}} dx$

    • $= \lim<em>{R \to \infty} \left[ 2\sqrt{x + 4} \right]</em>{1}^{R} = \lim_{R \to \infty} (2\sqrt{R + 4} - 2\sqrt{5}) = \infty$

  • Since the integral diverges, the series diverges.

P-Series

• Form: $\sum_{n=1}^{\infty} \frac{1}{n^{p}}$

• If p > 1, the series converges.

• If $p \leq 1$, the series diverges.

• Special Case

  • Harmonic Series: $\sum_{n=1}^{\infty} \frac{1}{n}$ (p=1), diverges.

• Example 1:

  • $\sum_{n=1}^{\infty} \frac{1}{n^{5}}$

  • p = 5 > 1

  • Therefore, the series converges.

• Example 2:

  • $\sum_{n=1}^{\infty} \frac{1}{n}$

  • $p = 1$

  • Therefore, the series diverges (Harmonic Series).

Direct Comparison Test

• If $0 \leq a<em>{n} \leq b</em>{n}$ for all $n$, and $\sum b<em>{n}$ converges, then $\sum a</em>{n}$ converges.

• If $0 \leq b<em>{n} \leq a</em>{n}$ for all $n$, and $\sum b<em>{n}$ diverges, then $\sum a</em>{n}$ diverges.

• Example 1:

  • $\sum_{n=5}^{\infty} \frac{1}{n - 4}$

  • Compare with $\sum_{n=5}^{\infty} \frac{1}{n}$

  • \frac{1}{n - 4} > \frac{1}{n}

  • Since $\sum<em>{n=5}^{\infty} \frac{1}{n}$ diverges (Harmonic Series), $\sum</em>{n=5}^{\infty} \frac{1}{n - 4}$ also diverges.

• Example 2:

  • $\sum_{n=1}^{\infty} \frac{1}{n^{3} + 5}$

  • Compare with $\sum_{n=1}^{\infty} \frac{1}{n^{3}}$

  • \frac{1}{n^{3} + 5} < \frac{1}{n^{3}}

  • Since $\sum<em>{n=1}^{\infty} \frac{1}{n^{3}}$ converges (P-Series with p = 3 > 1), $\sum</em>{n=1}^{\infty} \frac{1}{n^{3} + 5}$ also converges.

Limit Comparison Test

• If $\lim<em>{n \to \infty} \frac{a</em>{n}}{b<em>{n}} = L$, where 0 < L < \infty, then $\sum a</em>{n}$ and $\sum b_{n}$ either both converge or both diverge.

• Example 1:

  • $\sum_{n=1}^{\infty} \frac{1}{n^{2} - 4}$

  • Let $b_{n} = \frac{1}{n^{2}}$, which converges (P-Series with p = 2 > 1).

  • $\lim<em>{n \to \infty} \frac{\frac{1}{n^{2} - 4}}{\frac{1}{n^{2}}} = \lim</em>{n \to \infty} \frac{n^{2}}{n^{2} - 4} = 1$

  • Since the limit is a positive number, and $\sum b<em>{n}$ converges, $\sum a</em>{n}$ also converges.

• Example 2:

  • $\sum_{n=1}^{\infty} \frac{3n^{2} + 2n}{4n^{3} - 5}$

  • Let $b_{n} = \frac{1}{n}$, which diverges (Harmonic Series).

  • $\lim<em>{n \to \infty} \frac{\frac{3n^{2} + 2n}{4n^{3} - 5}}{\frac{1}{n}} = \lim</em>{n \to \infty} \frac{3n^{3} + 2n^{2}}{4n^{3} - 5} = \frac{3}{4}$

  • Since the limit is a positive number, and $\sum b<em>{n}$ diverges, $\sum a</em>{n}$ also diverges.

Alternating Series Test

• Consider an alternating series of the form: $\sum<em>{n=1}^{\infty} (-1)^{n} a</em>{n}$

• If

  • $\lim<em>{n \to \infty} a</em>{n} = 0$

  • $a_{n}$ is decreasing

• Then the alternating series converges.

• The a sub n only represents the non-alternating part.

• Conditional Convergence

Alternating Series Error Bound

• The error in approximating the sum of an alternating series by using a finite number of terms is less than or equal to the absolute value of the first unused term.

• If $s = \sum<em>{n=1}^{\infty} (-1)^{n} a</em>{n}$ and $s<em>{N} = \sum</em>{n=1}^{N} (-1)^{n} a<em>{n}$, then $\left| s - s</em>{N} \right| \leq a_{N+1}$

• Quick Example:

  • Given the series $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{2} + 1}$, approximate the sum using the first four terms. Find the error bound.

  • $Error \leq \left| \frac{(-1)^{5}}{5^{2} + 1} \right| = \frac{1}{26} \approx 0.038$

Ratio Test

• Let $L = \lim<em>{n \to \infty} \left| \frac{a</em>{n+1}}{a_{n}} \right|$

• If:

  • L < 1, the series converges.

  • L > 1, the series diverges.

  • $L = 1$, the test is inconclusive.

• Example 1:

  • $\sum_{n=1}^{\infty} \frac{2^{n}}{n!}$

  • $L = \lim<em>{n \to \infty} \left| \frac{\frac{2^{n+1}}{(n+1)!}}{\frac{2^{n}}{n!}} \right| = \lim</em>{n \to \infty} \frac{2^{n+1} n!}{2^{n} (n+1)!} = \lim_{n \to \infty} \frac{2}{n+1} = 0$

  • Since L = 0 < 1, the series converges.

• Example 2:

  • $\sum_{n=1}^{\infty} \frac{2^{n}}{n}$

  • $L = \lim<em>{n \to \infty} \left| \frac{\frac{2^{n+1}}{n+1}}{\frac{2^{n}}{n}} \right| = \lim</em>{n \to \infty} \frac{2^{n+1} n}{2^{n} (n+1)} = \lim_{n \to \infty} \frac{2n}{n+1} = 2$

  • Since L = 2 > 1, the series diverges.

Absolute vs. Conditional Convergence

• A series $\sum a<em>{n}$ converges absolutely if $\sum \left| a</em>{n} \right|$ converges.

• A series $\sum a<em>{n}$ converges conditionally if $\sum a</em>{n}$ converges, but $\sum \left| a_{n} \right|$ diverges.

• Example 1:

  • $\sum_{n=1}^{\infty} (-1)^{n} \frac{3}{n}$

  • $\sum<em>{n=1}^{\infty} \left| (-1)^{n} \frac{3}{n} \right| = \sum</em>{n=1}^{\infty} \frac{3}{n}$, which diverges (Harmonic Series).

  • However, $\sum_{n=1}^{\infty} (-1)^{n} \frac{3}{n}$ converges by the Alternating Series Test (terms tend to 0 and are decreasing).

  • Therefore, the series converges conditionally.

• Example 2:

  • $\sum_{n=1}^{\infty} \frac{2n + 3}{6n^{3} - 5n}$

  • $\sum<em>{n=1}^{\infty} \left| \frac{2n + 3}{6n^{3} - 5n} \right| = \sum</em>{n=1}^{\infty} \frac{2n + 3}{6n^{3} - 5n}$

  • Using the Limit Comparison Test with $b<em>{n} = \frac{1}{n^{2}}$, we find that $\sum</em>{n=1}^{\infty} \frac{2n + 3}{6n^{3} - 5n}$ converges.

  • Therefore, the series converges absolutely.