Convergence of Series with Positive Terms

For a sequence $\{a_n\}$, where $a_n$ is a non-negative real number for every $n$, the sequence of partial sums

$S_k = \sum_{n=0}^{k}a_n = a_0 + a_1 + \cdots + a_k$ 

is non-decreasing. Because the partial sum $S_k$ can only increase as $k$ becomes larger, the limit of the partial sum can either converge or diverge to $\infty$. Therefore, it follows that a series $\sum_{n=0}^{\infty} a_n$ with non-negative terms converges if and only if the sequence $S_k$ of partial sums is bounded.

Example 1

The series $\sum_{n \ge 1} \frac{1}{n^2}$ is convergent because of the inequality:

$\displaystyle{\frac1 {n^2} \le \frac{1}{n-1} - \frac{1}{n}, (n \ge 2)}$ 

and because:

 $\displaystyle{\sum_{n \ge 2} \left(\frac{1}{n-1} - \frac{1}{n} \right) =\left(1-\frac{1}{2}\right) + \left(\frac{1}{2}-\frac{1}{3}\right) + \left(\frac{1}{3}-\frac{1}{4}\right) + \cdots = 1}$

Example 2

Would the series

$\displaystyle{S = 1 + \frac{1}{2}+ \frac{1}{4}+ \frac{1}{8}+\cdots+ \frac{1}{2^n}+\cdots}$ 

converge? It is possible to "visualize" its convergence on the real number line? We can imagine a line of length $2$, with successive segments marked off of lengths $1$, $\frac{1}{2}$, $\frac{1}{4}$, etc. There is always room to mark the next segment, because the amount of line remaining is always the same as the last segment marked: when we have marked off $\frac{1}{2}$, we still have a piece of length $\frac{1}{2}$ unmarked, so we can certainly mark the next $\frac{1}{4}$. This argument does not prove that the sum is equal to $2$ (although it is), but it does prove that it is at most $2$. In other words, the series has an upper bound. Proving that the series is equal to $2$ requires only elementary algebra, however. If the series is denoted $S$, it can be seen that:

 $\displaystyle{\frac{S}{2} \,= \frac{1+ \frac{1}{2}+ \frac{1}{4}+ \frac{1}{8}+\cdots}{2} \\ \quad= \frac{1}{2}+ \frac{1}{4}+ \frac{1}{8}+ \frac{1}{16} +\cdots}$

Therefore:

$\displaystyle{S-\frac{S}{2} = 1\\S = 2}$

Geometric Sum

Visualization of the geometric sum in Example 2. The length of the line ($2$) can contain all the successive segments marked off of lengths $1$, $\frac{1}{2}$, $\frac{1}{4}$, etc.

For these specific examples, there are easy ways to check the convergence. However, it could be the case that there are no easy ways to check the convergence. For these general cases, we can experiment with several well-known convergence tests (such as ratio test, integral test, etc.). We will learn some of these tests in the following atoms.