
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 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 that describes the next term in the list.

### A recursive formula for $s_n$

As it turns out, there is always a recursive formula for $s_n$. For the sake of example, 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 $\sum _{k=1}^{\infty } \frac {1}{k}$, where $a_n = \frac {1}{n}$, the result read $s_n = s_{n-1} + \frac {1}{n}$. 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 $a_n = a\cdot r^n$.

This generates the ordered list

and we have a result for the limits of these types of sequences, which we now recall.

Note that the limit of this new sequence is exactly the sum of all of the terms in the old sequence! Let’s formalize the ideas in the last example with a definition.

The above definition really is assuring us that the symbols $\sum _{k=k_0} a_k$ and $\lim _{n \to \infty } s_n$ are exactly the same! However, this definition makes the content of the previous example more precise. The major idea here is that we have techniques that we can use to determine whether limits exist and can even find what those limits are sometimes. Since we are now able to recast the new question “Can I sum all of the terms in a sequence?” into the old question “Does a sequence have a limit?”, we can now utilize all of our previous techniques to analyze the sequence of partial sums.

Using our new terminology, what can we say about the series $\sum _{k=1}^\infty \left (\frac {1}{2}\right )^k$ from the previous example? The series convergesdiverges , and $\sum _{n=1}^\infty \left (\frac {1}{2}\right )^n = \answer [given]{1}$.

### Two special types of series

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.

We can now think of adding together 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 the limit in question, $\lim _{n \to \infty } r^{n+1}$ 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 .

In fact, if $r>1$, the $r^{n+1}$ is not bounded above. If $r=-1$, the $r^{n+1}$ is bounded, but the terms oscillate between $-1$ and $1$. If $r>1$, the terms $r^{n+1}$ both oscillate in sign and become arbitrarily large in magnitude.

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.