Telescoping Series

In this post, we will learn more about infinite series. We shall start with so-called telescoping series, that has a form of

$$ \begin{aligned} \sum_{k=1}^{\infty} (a_n - a_{n+1}) \end{aligned} $$
$$ \begin{aligned} \sum_{n=1}^{\infty} (a_n - a_{n+1}) \end{aligned} $$

Let us investigate this sum step by step. Observe that the partial sum $s_n$ can be written as

$$ \begin{aligned} s_n = a_1 - a_2 + a_2 - a_3 + \dots + a_n - a_{n+1} \end{aligned} $$
$$ \begin{aligned} s_n = a_1 - a_2 + a_2 - a_3 + \dots + a_n - a_{n+1} \end{aligned} $$

So we can just simplify this

$$ \begin{aligned} s_n = a_1 - a_{n+1} \end{aligned} $$
$$ \begin{aligned} s_n = a_1 - a_{n+1} \end{aligned} $$

We know that a series converges if and only if the partial sum of the series converges. Using this information we can create corollary: a telescoping series $\sum_{n=1}^{\infty} (a_n - a_{n+1})$ converges if and only if $\lim_{n \rightarrow \infty} a_{n+1} = \lim_{n \rightarrow \infty} a_n$ converges. We shall give an example.

$\textbf{\small Example:} $ Determine whether the following series is convergent or divergent.

$$ \begin{aligned} \sum_{n=1}^{\infty} \dfrac{1}{(\alpha + n)(\alpha + n + 1)} \end{aligned} $$
$$ \begin{aligned} \sum_{n=1}^{\infty} \dfrac{1}{(\alpha + n)(\alpha + n + 1)} \end{aligned} $$

$\textbf{\small Solution:} $ It seems irrelevant to the telescoping series, but we will divide this telescoping series term into two parts by using partial fraction decomposition.

$$ \begin{aligned} \dfrac{1}{(\alpha + n)(\alpha + n + 1)} = \dfrac{A}{\alpha + n} + \dfrac{B}{\alpha + n + 1} \end{aligned} $$
$$ \begin{aligned} \dfrac{1}{(\alpha + n)(\alpha + n + 1)} = \dfrac{A}{\alpha + n} + \dfrac{B}{\alpha + n + 1} \end{aligned} $$

What we see here is the decompisition of the telescoping series term with coeffecients $A$ and $B$. All we have to do is to determine the values for $A$ and $B$ respectively.

$$ \begin{aligned} &\dfrac{A\cdot (\alpha + n + 1)}{(\alpha + n)\cdot (\alpha + n + 1)} + \dfrac{B\cdot (\alpha + n)}{(\alpha + n)\cdot (\alpha + n + 1)} \\ =&\dfrac{A\cdot (\alpha + n + 1) + B\cdot (\alpha + n)}{(\alpha + n)\cdot (\alpha + n + 1)}\\ =&\dfrac{\alpha\cdot(A+B) + n\cdot(A+B) + A}{(\alpha + n)\cdot (\alpha + n + 1)} \end{aligned} $$
$$ \begin{aligned} &\dfrac{A\cdot (\alpha + n + 1)}{(\alpha + n)\cdot (\alpha + n + 1)} + \dfrac{B\cdot (\alpha + n)}{(\alpha + n)\cdot (\alpha + n + 1)} \\ =&\dfrac{A\cdot (\alpha + n + 1) + B\cdot (\alpha + n)}{(\alpha + n)\cdot (\alpha + n + 1)}\\ =&\dfrac{\alpha\cdot(A+B) + n\cdot(A+B) + A}{(\alpha + n)\cdot (\alpha + n + 1)} \end{aligned} $$

And equating this back into the telescoping series term we get

$$ \begin{aligned} &\dfrac{1}{(\alpha + n)(\alpha + n + 1)} \\ =&\dfrac{\alpha\cdot(A+B) + n\cdot(A+B) + A}{(\alpha + n)\cdot (\alpha + n + 1)} \end{aligned} $$
$$ \begin{aligned} &\dfrac{1}{(\alpha + n)(\alpha + n + 1)} \\ =&\dfrac{\alpha\cdot(A+B) + n\cdot(A+B) + A}{(\alpha + n)\cdot (\alpha + n + 1)} \end{aligned} $$

Basically,

$$ \begin{aligned} 1 = \alpha\cdot(A+B) + n\cdot(A+B) + A \end{aligned} $$
$$ \begin{aligned} 1 = \alpha\cdot(A+B) + n\cdot(A+B) + A \end{aligned} $$

It is obvious that $A+B=0$ and $A=1$. That is why $A = 1$ and $B = -1$. Putting everything back we can write the sum as

$$ \begin{aligned} \sum_{n=1}^{\infty} \dfrac{1}{\alpha + n} - \dfrac{1}{\alpha + n + 1} \end{aligned} $$
$$ \begin{aligned} \sum_{n=1}^{\infty} \dfrac{1}{\alpha + n} - \dfrac{1}{\alpha + n + 1} \end{aligned} $$
Using the corollary we have just derived, we know that if

$\lim_{n \rightarrow \infty} \frac{1}{\alpha + n}-\frac{1}{\alpha + n + 1} < \infty$, then $\sum_{n=1}^{\infty} \frac{1}{\alpha + n} - \frac{1}{\alpha + n + 1}$ converges, and vice versa. Obviously, $\lim_{n \rightarrow \infty} \frac{1}{\alpha + n}-\frac{1}{\alpha + n + 1} = 0$, thus the series converges.

Before moving into the more tests let us examine the algebraic properties of series.

Some algebra

Let $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} b_n$ be convergent series. Then the following properties are satisfied:

$$ \begin{align} &\sum_{n=1}^{\infty} a_n \pm b_n \\ &\sum_{n=1}^{\infty} c \cdot a_n = c \cdot \sum_{n=1}^{\infty} a_n \end{align} $$
$$ \begin{align} &\sum_{n=1}^{\infty} a_n \pm b_n \\ &\sum_{n=1}^{\infty} c \cdot a_n = c \cdot \sum_{n=1}^{\infty} a_n \end{align} $$

The proof is pretty straight-forward therefore I will not prove these here. I should define positive series and then move on to the new section.

$\textbf{\small Definition 1.20.:} $ (Positive Series) $\sum_{n=1}^{\infty} a_n$ is called positive if $a_n > 0$, for all $n \in \mathbb{N}$.

$\textbf{\small Theorem 1.14.:} $ (Convergence of Positive Series) The positive series $\sum_{n=1}^{\infty} a_n$ converges if and only if $(s_n)$ is bounded.

$\textbf{\small Proof:} $ $(\Rightarrow)$ Let $\sum_{n=1}^{\infty} a_n$ be a convergent series. Then the partial sum $(s_n)$ also converges. Trivially, $(s_n)$ is bounded. Remember that, every convergent sequence is bounded.

$(\Leftarrow)$ Let $(s_n)$ be bounded sequence, namely $s_n = a_1 + a_2 + a_3 + \dots + a_n$ is bounded. We also now, $a_n$ is increasing by assumption. Observe that,

$$ \begin{aligned} s_n-s_{n-1} = a_n > 0 \end{aligned} $$
$$ \begin{aligned} s_n-s_{n-1} = a_n > 0 \end{aligned} $$

This means that $(s_n)$ is also an increasing sequence, because

$$ \begin{aligned} s_n > s_{n-1} \end{aligned} $$
$$ \begin{aligned} s_n > s_{n-1} \end{aligned} $$

Thus the partial sum $(s_n)$ is monotone increasing bounded sequence, by monotone convergence theorem we can say $(s_n)$ converges.

$\therefore$ by definition $\sum_{n=1}^{\infty} a_n$ converges.

From now on we will give techniques/tests for examining the convergence of series.

Tests

Direct Comparison Test

$\textbf{\small Theorem 1.14.:} $ (Direct Comparison Theorem) Let $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} b_n$ be positive series. If $0< a_n < b_n$ and $\sum_{n=1}^{\infty} b_n$ converges, then $\sum_{n=1}^{\infty} a_n$ converges. Similarly, if $\sum_{n=1}^{\infty} a_n$ diverges, then $\sum_{n=1}^{\infty} b_n$ diverges.

$\textbf{\small Proof:} $ Let $0 < a_n < b_n$ and $\sum_{n=1}^{\infty} b_n$ converges. We will prove that $\sum_{n=1}^{\infty} a_n$ also converges. Let the partial sum of the series $\sum_{n=1}^{\infty} b_n$ be $t_n = b_1 + b_2 + b_3 + \dots + b_n$.

By definition, we know $\lim_{n \rightarrow \infty} t_n$ converges, say $t$. Observe that, for the partial sum $(s_n)$ of the $\sum_{n=1}^{\infty} a_n$, the following inequality holds; $(s_n) < (t_n) < t$ by the assumption of the theorem. Also, for all $n>1$ we have $s_n - s_{n-1} = a_n > 0$, means that $(s_n)$ is monotone increasing.

Notice that, we have everything to devise monotone convergence theorem, because $(s_n)$ is bounded above by $t$ and is also monotone increasing sequence. Hence, we conclude that $(s_n)$ converges, so is $\sum_{n=1}^{\infty} a_n$ converges.

Now, let us prove the other statement. Let $0 < a_n < b_n$ and $\sum_{n=1}^{\infty} a_n$ diverges. Suppose that $\sum_{n=1}^{\infty} b_n$ converges. Then using first part $\sum_{n=1}^{\infty} a_n$ converges, which contradicts the assumption that $\sum_{n=1}^{\infty} a_n$ diverges. Therefore, $\sum_{n=1}^{\infty} b_n$ diverges.

To comprehend the theorem more, we shall examine the following example

$\textbf{\small Example:} $ Determine whether the following series converges

$$ \begin{aligned} \sum_{n=1}^{\infty} \dfrac{1}{n^3} \end{aligned} $$
$$ \begin{aligned} \sum_{n=1}^{\infty} \dfrac{1}{n^3} \end{aligned} $$

$\textbf{\small Solution:} $ Observe that $0 < \frac{1}{n^3} < \frac{1}{n^2}$. Also, we know by previous posts that $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges. Devising the direct comparison test we conclude that $\sum_{n=1}^{\infty} \frac{1}{n^3}$ converges.

Now, let let us expand this idea.

$p$-series

$\textbf{\small Theorem 1.15.:} $ ($p$-series) The series

$$ \begin{aligned} \sum_{n=1}^{\infty} \dfrac{1}{n^p} \end{aligned} $$
$$ \begin{aligned} \sum_{n=1}^{\infty} \dfrac{1}{n^p} \end{aligned} $$

is called the $p$-series. If $p > 1$ then the sum is finite and for $p \leq 1$ is infinite. For $p=1$ we have harmonic series.

The sum $\sum_{n=1}^{\infty} \frac{1}{n^p}$, when $p>1$ is called Riemann $\zeta(p)$ function and it contains deep, mystic and unsolved questions.

$\textbf{\small Proof:} $ If $p \leq 0$, then $\lim_{n \rightarrow \infty} \frac{1}{n^p} \neq 0$, then the series $\sum_{n=1}^{\infty} \frac{1}{n^p}$ diverges by $n$-th term test. If $p = 0$, then we have so-called harmonic series.

If $0 < p < 1$, then $n^p < n$ and by direct comparison test the series $\sum_{n=1}^{\infty} \frac{1}{n^p}$ diverges ($0 < \frac{1}{n} < \frac{1}{n^p}$ and $\sum_{n=1}^{\infty} \frac{1}{n}$ diverges).

If $p > 1$, then

$$ \begin{aligned} \sum_{n=1}^{\infty} \dfrac{1}{n^p} &= 1 + \dfrac{1}{2^p} + \dfrac{1}{3^p} + \dots + \dfrac{1}{n^p} + \dots \\ &< 1 + \dfrac{2}{2^p} + \dfrac{4}{4^p} + \dots + \dfrac{n}{n^{2p}} + \dots \\ &= \sum_{n=1}^{\infty} \left(\dfrac{1}{2^{p-1}}\right)^n \end{aligned} $$
$$ \begin{aligned} \sum_{n=1}^{\infty} \dfrac{1}{n^p} &= 1 + \dfrac{1}{2^p} + \dfrac{1}{3^p} + \dots + \dfrac{1}{n^p} + \dots \\ &< 1 + \dfrac{2}{2^p} + \dfrac{4}{4^p} + \dots + \dfrac{n}{n^{2p}} + \dots \\ &= \sum_{n=1}^{\infty} \left(\dfrac{1}{2^{p-1}}\right)^n \end{aligned} $$

The idea we used here is as follows:

$$ \begin{aligned} &\dfrac{1}{2^p} + \dfrac{1}{3^p} < \dfrac{1}{2^p} + \dfrac{1}{2^p} = \dfrac{2}{2^p}\\ &\dfrac{1}{4^p} + \dots + \dfrac{1}{7^p} < \dfrac{1}{4^p} + \dots +\dfrac{1}{4^p} = \dfrac{4}{4^p}\\ \end{aligned} $$
$$ \begin{aligned} &\dfrac{1}{2^p} + \dfrac{1}{3^p} < \dfrac{1}{2^p} + \dfrac{1}{2^p} = \dfrac{2}{2^p}\\ &\dfrac{1}{4^p} + \dots + \dfrac{1}{7^p} < \dfrac{1}{4^p} + \dots +\dfrac{1}{4^p} = \dfrac{4}{4^p}\\ \end{aligned} $$

and so on. Notice that, the geometric series $\sum_{n=1}^{\infty} \left(\frac{1}{2^{p-1}}\right)^n$ converges with $r = \frac{1}{2^{p-1}} < 1$ (look at geometric series). Also, $0 < \frac{1}{n^p} < \left(\frac{1}{2^{p-1}}\right)^n$ and since $\sum_{n=1}^{\infty} \left(\frac{1}{2^{p-1}}\right)^n$ converges. We conclude that $\sum_{n=1}^{\infty}\frac{1}{n^{p}}$ converges by direct comparison test.

There are some famous examples of $p$-series:

$$ \begin{aligned} &\zeta(2) = \sum_{n=1}^{\infty}\frac{1}{n^{2}} = \dfrac{\pi^2}{6}\\ &\zeta(3) = \text{unknown} \\ &\zeta(4) = \sum_{n=1}^{\infty}\frac{1}{n^{4}} = \dfrac{\pi^4}{90}\\ \end{aligned} $$
$$ \begin{aligned} &\zeta(2) = \sum_{n=1}^{\infty}\frac{1}{n^{2}} = \dfrac{\pi^2}{6}\\ &\zeta(3) = \text{unknown} \\ &\zeta(4) = \sum_{n=1}^{\infty}\frac{1}{n^{4}} = \dfrac{\pi^4}{90}\\ \end{aligned} $$

Limit Comparison Test

$\textbf{\small Theorem 1.16.:} $ Let $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} b_n$ be positive series. If

$$ \begin{aligned} \lim_{n \rightarrow \infty} \dfrac{a_n}{b_n} = c \end{aligned} $$
$$ \begin{aligned} \lim_{n \rightarrow \infty} \dfrac{a_n}{b_n} = c \end{aligned} $$

where $0 < c < \infty$, then either both series converge or diverge.

$\textbf{\small Proof:} $ Suppose $\lim_{n \rightarrow \infty} \dfrac{a_n}{b_n} = c$ where $0 < c < \infty$. Then,

$$ \begin{aligned} \left| \dfrac{a_n}{b_n} - c \right| &< \epsilon \\ -\epsilon < \dfrac{a_n}{b_n} - c &< \epsilon \\ c - \epsilon < \dfrac{a_n}{b_n} &< c + \epsilon \\ b_n\cdot (c - \epsilon) < a_n &< b_n \cdot(c + \epsilon) \end{aligned} $$
$$ \begin{aligned} \left| \dfrac{a_n}{b_n} - c \right| &< \epsilon \\ -\epsilon < \dfrac{a_n}{b_n} - c &< \epsilon \\ c - \epsilon < \dfrac{a_n}{b_n} &< c + \epsilon \\ 0 < b_n\cdot (c - \epsilon) < a_n &< b_n \cdot(c + \epsilon) \end{aligned} $$

Take,

$$ \begin{aligned} 0 < b_n <\left(\dfrac{1}{c-\epsilon}\right)\cdot a_n \end{aligned} $$
$$ \begin{aligned} 0 < b_n <\left(\dfrac{1}{c-\epsilon}\right)\cdot a_n \end{aligned} $$

if $\sum_{n=1}^{\infty} a_n$ converges, then $\sum_{n=1}^{\infty} b_n$ converges by direct comparison test. Similary take,

$$ \begin{aligned} 0 < a_n <(\epsilon + c)\cdot b_n \end{aligned} $$
$$ \begin{aligned} 0 < a_n <(\epsilon + c)\cdot b_n \end{aligned} $$

if $\sum_{n=1}^{\infty} a_n$ diverges, then $\sum_{n=1}^{\infty} b_n$ also diverges by direct comparison test.

The following example will clearly demonstrate how to use the theorem.

$\textbf{\small Example:} $ Determine whether the following series converges or diverges

$$ \begin{aligned} \sum_{n=1}^{\infty} \sin\left(\dfrac{1}{n}\right) \end{aligned} $$
$$ \begin{aligned} \sum_{n=1}^{\infty} \sin\left(\dfrac{1}{n}\right) \end{aligned} $$

$\textbf{\small Solution:} $ First of all, we should show that $\sin\left(\dfrac{1}{n}\right)$ is positive for all $n > 0$. Notice that, $0 < \frac{1}{n} \leq 1$ for $n > 0$. And thus,

$$ \begin{aligned} \sin\left(0\right) < \sin\left(\dfrac{1}{n}\right) \leq \sin\left(1\right) \end{aligned} $$
$$ \begin{aligned} \sin\left(0\right) < \sin\left(\dfrac{1}{n}\right) \leq \sin\left(1\right) \end{aligned} $$

Means that $0 < \sin\left(\frac{1}{n}\right)$ hence $\sum_{n=1}^{\infty} \sin\left(\frac{1}{n}\right)$ is positive series. Now consider the series $\sum_{n=1}^{\infty} \frac{1}{n}$ and observe the limit

$$ \begin{aligned} \lim_{n\rightarrow \infty} \dfrac{\sin\left(\dfrac{1}{n}\right)}{\dfrac{1}{n}} = 1 \end{aligned} $$
$$ \begin{aligned} \lim_{n\rightarrow \infty} \dfrac{\sin\left(\dfrac{1}{n}\right)}{\dfrac{1}{n}} = 1 \end{aligned} $$

This means $\sum_{n=1}^{\infty}\sin\left(\frac{1}{n}\right)$ behaves as same as $\sum_{n=1}^{\infty} \frac{1}{n}$. In conclusion $\sum_{n=1}^{\infty}\sin\left(\frac{1}{n}\right)$ diverges by limit comparison test.

In the next section we will discuss and prove other tests including D’Alembert Ratio Test, Root Test, Raabe’s Test and Cauchy Condensation Test. Goodbye 👋.