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We discuss convergence results for geometric series and telescoping series.

Suppose that we have a series $\sum _{k=k_0}^{\infty } a_k$ and have to determine whether it converges or diverges. To answer this question, we define a new sequence $\{s_n\}_{n=k_0}$ where $s_n = \sum _{k=k_0}^{n}$ for all $n \geq k_0$. We saw previously that
• the series $\sum _{k=k_0}^\infty a_k$ converges if and only if $\lim _{n\to \infty } s_n$ exists.
• the series $\sum _{k=k_0}^\infty a_k$ diverges if and only if $\lim _{n\to \infty } s_n$ does not exist.

The definitions above give us a way to determine whether a given series converges. In fact, to determine whether $\sum _{k=k_0}^{\infty } a_k$ converges, we can do the following.

• Consider the associated sequence $\{s_n\}$ of partial sums.
• Try to find an explicit formula for the term $s_n$. If you can find such a formula, analyze $\lim _{n \to \infty } s_n$.
• If the limit exists, $\sum _{k=k_0}^{\infty } a_k$ converges, and if we can determine that $\lim _{n \to \infty } s_n =L$, then $\sum _{k=k_0} a_k=L$.
• If $\lim _{n \to \infty } s_n$ does not exist, then $\sum _{k=k_0} a_k$ diverges.
• If an explicit formula for $s_n$ cannot be found, further analysis is needed. We’ll expound on this in later sections.

### A recursive formula for $s_n$

The most straightforward way to determine whether $\lim _{n \to \infty } s_n$ exists is to have an explicit formula for the $n$-th term $s_n$. Note that this is not an easy task; for example, can you find a formula for $s_n$ for the series $\sum _{k=1}^{\infty } \frac {1}{k}$? It’s not too hard to write out the first several terms in the sequence $\{s_n\}_{n=1}$, but try to find an explicit formula for $s_n$!

As it turns out, there is always a recursive formula for $s_n$, and this will play an important role in later sections. Suppose that we want to consider $\sum _{k=1}^{\infty } a_k$. Let’s write out the formula for $s_n$.

We can make an observation by considering $s_{n-1}$ in a similar way.

Now returning to our expression for $s_n$, we can make an observation.

We thus have the formula

If we apply this to the series $\sum _{k=1}^{\infty } \frac {1}{k}$, we have $s_n = \sum _{k=1}^n \frac {1}{k}$ and $a_n = \frac {1}{n}$. The recursive formula reads

This does not help us analyze whether $\lim _{n \to \infty } s_n$ actually exists. Sometimes, however, we can find an explicit formula for $s_n$, and we study two special types of series for which this is possible.

### Geometric series

Recall that a geometric sequence is a sequence for which the ratio of successive terms is constant. If $\{a_n\}_{n=n_0}$ is such a sequence, then there are constants $a \ne 0$ and $r$ for which $a_n = a\cdot r^n$.

We thus represent this sequence by the ordered list

and we have a result that characterizes the behavior of this type of sequence, which we recall now.

We can now ask when we are able to sum the terms of a geometric sequence.

Before exploring when such a series converges, note that sometimes, some preliminary algebra is necessary to recognize a series as geometric.

We can now try to determine when adding together the terms in such a series is possible; that is, we can explore for which values of $a$ and $r$ the series $\sum _{k=n_0}^{\infty } a_k$ converges.

From our work above, we see that the $n$-th partial sum of the geometric series $a_n = r^n$ is We now have an explicit formula so we can determine for which values of $r$ the limit $\lim _{n \to \infty } s_n$ exists. First, note that by using the limit laws,

The existence of $\lim _{n \to \infty } s_n$is thus entirely determined by whether $\lim _{n \to \infty } r^{n+1}$ exists, and this limit is the limit of a geometric sequence! In fact,

• if $-1, then $\lim _{n \to \infty } r^{n+1}$ existsdoes not exist .
• if $r>1$ or $r\le -1$, then $\lim _{n \to \infty } r^{n+1}$ existsdoes not exist .

The above formula covers every case except when $r= 1$, but notice that so if $r=1$, $s_n = \answer [given]{n+1}$ and $\lim _{n \to \infty } s_n = \infty$, so $\sum _{k=0}^{\infty } 1$ diverges.

When $-1, note $\lim _{n \to \infty } r^{n+1}=0$, so in this case,

By noting that $\sum _{k=0}^n ar^k = a \sum _{k=0}^n r^k$, we can combine this observation with the above argument and write the result in a theorem.

There is a useful trick that allows us to find the sum of a convergent geometric series when the lower index does not start at $0$.

We can easily generalize this example and doing so allows us to write down a more comprehensive theorem about geometric series.

Now, try some questions to check your understanding of the above material.

Which of the following series converge?
$\sum _{k=0}^\infty \left (\frac {3}{2}\right )^k$ $\sum _{k=0}^\infty \left (\frac {-2}{3}\right )^k$ $\sum _{k=9}^\infty \left (\frac {1}{7}\right )^k$ $\sum _{k=1}^\infty (-1)^k$ $\sum _{k=-9}^\infty \left (\frac {1}{2}\right )^k$
The initial index doesn’t matter as far as convergence is concerned, it is the “tail” of the sequence that determines convergence.
Determine if the series $\sum _{k=2}^{\infty } 2^{3-2k}$ converges or diverges. If it converges, give the value to which it converges.

### Telescoping series

A second type of series for which we can find an explicit formula for $s_n$ are “telescoping series”. Rather than try to give a formal definition, we think of telescoping series are infinite sums for which the required addition required to find a formula for $s_n$ can be done so many of the intermediate terms naturally cancel. An example will make this point more clear.

We’ve just seen an example of a telescoping series. Informally, a telescoping series is one in which the partial sums reduce to just a finite sum of terms. In the last example, the partial sum $s_n$ only was the sum of two nonzero terms:

### Summary

Now that we have seen two special types of series for which we can find an explicit formula for the $n$-th term in the sequence of partial sums, it helps to summarize the logic that we employed.

• Consider the associated sequence $\{s_n\}$ of partial sums.
• Try to find an explicit formula for the term $s_n$. If you can find such a formula, analyze $\lim _{n \to \infty s_n}$.
• If the limit exists, $\sum _{k=k_0}^{\infty } a_k$ converges, and if we can determine that $\lim _{n \to \infty } s_n =L$, then $\sum _{k=k_0} a_k=L$.
• If $\lim _{n \to \infty } s_n$ does not exist, then $\sum _{k=k_0} a_k$ diverges.