Mastering Mathematics Smartly
by Wee Wen Shih

A unique self-help website that provides comprehensive coverage of mathematics at A-level & beyond, written in a student-friendly style.

Convergence of series

We will study methods to determine if a series converges. Our discussion will focus on series in which all the terms are positive.

First, we look at the Comparison Test.

Result 1: Let \sum a_n and \sum b_n be two positive series such that there is a positive integer m for which a_k \leq b_k for all integers k \geq m. Then:

(1) If \sum b_n converges, so does \sum a_n;

(2) If \sum a_n diverges, so does \sum b_n.

Before we look at examples, we quote an extremely useful result which is often used with the comparison test.

Result 2: Consider the series \sum \frac{1}{n^p}, where p is a constant. This is called a p-series. Then:

(a) If p > 1, the series \sum \frac{1}{n^p} converges.

(b) If p \leq 1, the series \sum \frac{1}{n^p} diverges.

Example 1: Determine whether \sum \frac{1}{1 + ln\:n} converges.

Let a_n = \frac{1}{n} and b_n = \frac{1}{1 + ln\:n}. Now a_n \leq b_n for n \geq 1. By result 1(2), if \sum \frac{1}{n} diverges [ apply result 2(b) ], so does \sum \frac{1}{1 + ln\:n}.


Example 2: Determine whether \frac{1}{2} + \frac{1}{2.2^2} + \frac{1}{3.2^3} + \frac{1}{4.2^4} + ...  converges.

Let a_n = \frac{1}{n2^n} and b_n = \frac{1}{2^n}. Now a_n \leq b_n for n \geq 1 and \sum \frac{1}{2^n} is a convergent geometric series. By result 1(1), \frac{1}{2} + \frac{1}{2.2^2} + \frac{1}{3.2^3} + \frac{1}{4.2^4} + ...  converges.


Next, we consider the Limit Comparison Test.

Result 3: Let \sum a_n and \sum b_n be two positive series such that L = \lim_{n \rightarrow +\infty} \frac{a_n}{b_n} exists and 0 < L < +\infty. Then \sum a_n converges/diverges if and only if \sum b_n converges/diverges. Again, the results involving p-series can be applied.

Example 3: Determine whether 1 + \frac{3}{5} + \frac{4}{10} + \frac{5}{17} + ...  converges.

The infinite series can be expressed as \sum \frac{n+1}{n^2+1}. Let a_n = \frac{n+1}{n^2+1} and b_n = \frac{1}{n}. The method to obtain b_n is to consider the leading terms in the expression of a_n, i.e. \frac{n}{n^2} = \frac{1}{n}.

Now

\\\lim_{n \rightarrow +\infty} \left[ \frac{a_n}{b_n} \right] \\=\lim_{n \rightarrow +\infty} \left[ \frac{n(n+1)}{n^2+1} \right] \\=\lim_{n \rightarrow +\infty} \left[ 1 + \frac{n-1}{n^2+1} \right] \\=1
 
and since \sum \frac{1}{n} diverges [ apply result 2(b) ], \sum \frac{n+1}{n^2+1} diverges by result 3.

Example 4: Determine whether \sum \frac{2n}{(n+1)(n+2)(n+3)} converges.

Let a_n = \frac{2n}{(n+1)(n+2)(n+3)} and b_n = \frac{1}{n^2}. Again, we are able to obtain b_n by considering the leading terms in the expression of a_n, i.e. \frac{n}{n^3} = \frac{1}{n^2}.

Now
\\\lim_{n \rightarrow +\infty} \left[\frac{a_n}{b_n}\right] \\=\lim_{n \rightarrow +\infty} \left[\frac{2n^3}{(n+1)(n+2)(n+3)}\right] \\=\lim_{n \rightarrow +\infty} \left[\frac{2n^3}{(n+1)(n+2)(n+3)} \times \frac{\frac{1}{n^3}}{\frac{1}{n^3}} \right] \\=\lim_{n \rightarrow +\infty} \left[\frac{2}{\left(1 + \frac{1}{n}\right)\left(1 + \frac{2}{n}\right)\left(1 + \frac{3}{n}\right)} \right] \\=2

and since \sum \frac{1}{n^2} converges [ apply result 2(a) ], \sum \frac{2n}{(n+1)(n+2)(n+3)} converges by result 3.

Practice:
Q1 Determine, by the comparison test, whether \sum 2^{-n}sin\left(n\right) converges.

Q2 Determine, by the limit comparison test, whether \sum \frac{n+5}{n^3-2n+3} converges.