One of the most important things you will need to learn in this section of the course is a list of standard examples of convergent and divergent series. (The reasons for this will be clear when we get on to discussing the comparison test for convergence.) And by far the most important examples are the series $\sum _{n=1}^{\infty}{x}^{n}$ for various fixed positive values of $x$ and $\sum _{n=1}^{\infty}{n}^{-p}$ for various fixed positive values of $p$. We consider these here.

**Example 2.1**** (geometric series)**

The series $\sum _{n=0}^{\infty}{x}^{n}$ converges to $\frac{1}{1-x}$ if $0<x<1$ and diverges if $1\le x$.

**Proof**

If $0<x<1$ we consider the sequence of partial sums as usual, defined as ${s}_{n}=\sum _{k=0}^{n}{x}^{k}$. By multiplying through by $x$ we get $x\cdot {s}_{n}=\sum _{k=0}^{n}x\cdot {x}^{k}=\sum _{k=1}^{n+1}{x}^{k}$. Subtracting these two equations we get $(1-x){s}_{n}={x}^{0}-{x}^{n+1}$ so ${s}_{n}=\frac{{x}^{0}-{x}^{n+1}}{1-x}$. If $0<x<1$ then the sequence ${x}^{n+1}$ converges to $0$ as $n\to \infty $, as proved elsewhere, and of course $1-x\ne 0$, so by continuity ${s}_{n}\to \frac{{x}^{0}-0}{1-x}=\frac{1}{1-x}$, as required.

If $x\ge 1$ then ${x}^{n}\ge 1$ for all $n$ so ${x}^{n}$ does not define a null sequence and the series diverges by the null sequence test.

If you prefer your series to start at $n=1$ you need to repeat the proof to show that the series $\sum _{n=1}^{\infty}{x}^{n}$ converges to $\frac{x}{1-x}$. I find the case starting at $n=0$ easier to remember.

In the next group of examples, we are not so lucky as to be able to give a nice formula for the limit. We can still analyse convergence using monotonicity, however. Recall that the exponent $p$ may be rational, in which case it is interpreted via some $s$th root.

**Example 3.1**

The series $\sum _{n=1}^{\infty}{n}^{-1}$ diverges.

**Proof**

We have the following inequalities

and so on:

Putting these together we have, for $m\in \mathbb{N}$,

which shows that the sequence of partial sums ${s}_{n}=\sum _{k=1}^{n}\frac{1}{n}$ is unbounded and hence not convergent.

The series $\sum _{n=1}^{\infty}{n}^{-1}$ comes up a lot, and is called the harmonic series.

**Example 3.2**

For $p\in \mathbb{Q}$, the series $\sum _{n=1}^{\infty}{n}^{-p}$ converges if $p>1$.

**Proof**

Suppose $p>1$. We group terms together in an apparently rather similar sort of way as in the last example, but this time getting an upper bound for the partial sums.

and so on. This gives

Put $q=p-1$ where $q\in \mathbb{Q}$ is positive. Then

so the sequence of partial sums, $\left({s}_{n}\right)$, is bounded. But as each term in the series is positive, $\left({s}_{n}\right)$ is monotonic and hence convergent.

The series $\sum _{n=1}^{\infty}{n}^{-p}$ also diverges if $0<p<1$. We will prove this later as an application of the comparison test.

Note that, in obvious contrast with $\sum _{n=1}^{\infty}{x}^{n}$ where we were able to state the sum of the series when it has one, the numerical values of $\sum _{n=1}^{\infty}{n}^{-p}$ for $p>1$ are rather difficult to calculate and we resorted to showing the limit exists without actually finding this limit. This latter situation is rather typical: series are often difficult to compute exactly.

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