Convergence Tests for Positive Series

Theory

When a series has positive terms (a_n>0) and the partial sums are difficult to compute directly, the following tests are used.

1. Integral Test:
   If $a_n=f(n)$ where $f$ is positive, continuous, and decreasing, then

$$\sum a_n \text{ converges } \Longleftrightarrow \int_1^\infty f(x)\,dx \text{ converges}.$$

2. Comparison Test:
   If $0 \le a_n \le b_n$:

• $\sum b_n$ converges $\Rightarrow \sum a_n$ converges
• $\sum b_n$ diverges $\Rightarrow \sum a_n$ diverges

3. Limit Comparison Test:
   If

\lim_{n\to\infty} \frac{a_n}{b_n} = c > 0,

then $\sum a_n$ and $\sum b_n$ either both converge or both diverge. 4. Ratio Test:
Let

$$L = \lim_{n\to\infty} \frac{a_{n+1}}{a_n}.$$

If L<1, the series converges; if L>1, it diverges. 5. Root Test:
Let

$$R = \lim_{n\to\infty} \sqrt[n]{a_n}.$$

If R<1, the series converges; if R>1, it diverges.

Beginner Pattern

1. Identify the test
2. Simplify the expression
3. Take the limit or evaluate the integral
4. State convergence or divergence

Exercises

Exercise 1

Question

Integral Test Eg.a:

$$\sum_{n=1}^{\infty} ne^{-n}$$

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Solution

Let $f(x)=xe^{-x}$.

$$\int_1^\infty xe^{-x}\,dx = \left[-(x+1)e^{-x}\right]_1^\infty = \frac{2}{e}.$$

Conclusion: The series converges.

Exercise 2

Question

Integral Test Eg.b:

$$\sum_{n=1}^{\infty} \frac{1}{(n+1)\sqrt{\ln(n+1)}}$$

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Solution

Compare with

$$\int \frac{1}{x\sqrt{\ln x}}\,dx = 2\sqrt{\ln x}.$$

As $x\to\infty$, the integral diverges.

Conclusion: The series diverges.

Exercise 3

Question

Comparison Test Eg.a:

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

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Solution

\frac{1}{2^{n-1}+1} < \frac{1}{2^{n-1}}.

Since $\sum \frac{1}{2^{n-1}}$ converges,

Conclusion: The series converges.

Exercise 4

Question

Comparison Test Eg.b:

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

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Solution

$$\frac{1}{n^n} \le \frac{1}{2^n} \quad (n\ge2).$$

Since $\sum \frac{1}{2^n}$ converges,

Conclusion: The series converges.

Exercise 5

Question

Comparison Test Eg.c:

$$\sum_{n=1}^{\infty} \frac{1}{3^n-\cos n}$$

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Solution

$$3^n-\cos n \ge 3^n-1, \quad \frac{1}{3^n-\cos n} \le \frac{1}{3^n-1}.$$

Since $\sum \frac{1}{3^n}$ converges,

Conclusion: The series converges.

Exercise 6

Question

Limit Comparison Eg.a:

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

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Solution

Compare with $\sum \frac{1}{n^2}$.

$$\lim_{n\to\infty} \frac{n/(4n^3-2)}{1/n^2} = \frac14.$$

Conclusion: The series converges.

Exercise 7

Question

Limit Comparison Eg.b:

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

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Solution

Compare with $\sum \frac{1}{\sqrt{n}}$.

$$\lim_{n\to\infty} \frac{\ln n/\sqrt{n}}{1/\sqrt{n}} = \infty.$$

Conclusion: The series diverges.

Exercise 8

Question

Limit Comparison Eg.c:

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

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Solution

Compare with $\sum \frac{1}{n^{3/2}}$.

$$\lim_{n\to\infty} \frac{\ln n/n^2}{1/n^{3/2}} = 0.$$

Conclusion: The series converges.

Exercise 9

Question

Ratio Test Eg.a:

$$\sum_{n=1}^{\infty} \frac{3^n}{n!}$$

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Solution

$$\frac{a_{n+1}}{a_n}=\frac{3}{n+1}\to0.$$

Conclusion: The series converges.

Exercise 10

Question

Ratio Test Eg.b:

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

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Solution

$$\frac{a_{n+1}}{a_n}=\frac{(2n+2)(2n+1)}{(n+1)^2}\to4.$$

Conclusion: The series diverges.

Exercise 11

Question

Root Test Eg:

$$\sum_{n=1}^{\infty} \frac{1}{[\ln(n+1)]^n}$$

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Solution

$$\sqrt[n]{a_n}=\frac{1}{\ln(n+1)}\to0.$$

Conclusion: The series converges.