Telescoping Series Examples 1

# Telescoping Series Examples 1

We will now look at some more examples of evaluating telescoping series. Be sure to review the Telescoping Series page before continuing forward.

More examples can be found on the Telescoping Series Examples 2 page.

## Example 1

Determine whether the series $\sum_{n=1}^{\infty} \frac{1}{(2n - 1)(2n + 1)}$ is convergent or divergent. If this series is convergent find its sum.

Let's first expand this series as follows:

(1)
\begin{align} \quad \sum_{n=1}^{\infty} \frac{1}{(2n - 1)(2n + 1)} = \frac{1}{1 \times 3} + \frac{1}{3 \times 5} + \frac{1}{5 \times 7} + ... \end{align}

What we're going to want to is apply the techniques of partial fractions to rewrite the general term $a_n = \frac{1}{(2n -1)(2n+1)}$. So there exists real numbers $A$ and $B$ such that:

(2)
\begin{align} \frac{1}{(2n-1)(2n+1)} = \frac{A}{(2n-1)} + \frac{B}{(2n+1)} = \frac{A(2n+1) + B(2n-1)}{(2n-1)(2n+1)} \end{align}

Therefore $1 = A(2n+1) + B(2n-1)$. This equation must be true for all $n \in \mathbb{N}$. Even better, this equation is true for all $n \in \mathbb{R}$. If we let $n = \frac{1}{2}$ we get that $1 = 2A + 0B$ and so $A = \frac{1}{2}$. If we let $n = -\frac{1}{2}$ we get that $1 = 0A - 2B$ and so $B = \frac{-1}{2}$. So then:

(3)
\begin{align} \frac{1}{(2n-1)(2n+1)} = \frac{1}{4n - 2} -\frac{1}{4n + 2} \end{align}

We thus get that:

(4)
\begin{align} \quad \sum_{n=1}^{\infty} \frac{1}{(2n - 1)(2n + 1)} = \sum_{n=1}^{\infty} \left ( \frac{1}{4n - 2} -\frac{1}{4n + 2} \right ) \end{align}

Now consider the $n^{\mathrm{th}}$ partial sum:

(5)
\begin{align} \quad \quad s_n = \sum_{i=1}^{n} \left ( \frac{1}{4i - 2} -\frac{1}{4i + 2} \right ) = \left ( \frac{1}{2} -\frac{1}{6} \right) + \left ( \frac{1}{6} - \frac{1}{10} \right ) + \left ( \frac{1}{10} -\frac{1}{14} \right) + ... + \left ( \frac{1}{4n - 2} -\frac{1}{4n + 2} \right ) \\ \quad s_n = \frac{1}{2} - \frac{1}{4n + 2} \end{align}

Therefore the sequence of partial sums $\{ s_n \} = \{ \frac{1}{2} - \frac{1}{4n + 2} \}$ and thus:

(6)
\begin{align} \lim_{n \to \infty} s_n = \lim_{n \to \infty} \left ( \frac{1}{2} - \frac{1}{4n + 2} \right ) = \frac{1}{2} \end{align}

Therefore the series is convergent and $\sum_{n=1}^{\infty} \frac{1}{(2n-1)(2n+2)} = \frac{1}{2}$.

## Example 2

Determine whether the series $\sum_{n=1}^{\infty} \frac{3}{n^2 + 3n + 2}$ is convergent or divergent. If this series is convergent find its sum.

We are going to first want to split the general term $a_n = \frac{3}{n^2 + 3n + 2}$. We first factor the denominator to get that $a_n = \frac{3}{(n+1)(n+2)}$. We thus want to find real numbers $A$ and $B$ such that:

(7)
\begin{align} \frac{3}{(n+1)(n+2)} = \frac{A}{(n+1)} + \frac{B}{(n+2)} = \frac{A(n+2) + B(n+1)}{(n+1)(n+2)} \end{align}

Therefore we have that $3 = A(n+2) + B(n+1)$. If $n = -2$ then $3 = 0A -B$ and so $B = -3$. If $n = -1$ then $3 = A + 0B$ and so $A = 3$. Therefore:

(8)
\begin{align} \frac{3}{(n+1)(n+2)} = \frac{3}{(n+1)} - \frac{3}{(n+2)} \end{align}

And so it follows that:

(9)
\begin{align} \quad \sum_{n=1}^{\infty} \frac{1}{n^2 + 3n + 2} = \sum_{n=1}^{\infty} \left ( \frac{3}{(n+1)} - \frac{3}{(n+2)} \right ) \end{align}

Now consider the $n^{\mathrm{th}}$ partial sum:

(10)
\begin{align} \quad \quad s_n = \sum_{i=1}^{n} \left ( \frac{3}{(i+1)} - \frac{3}{(i+2)} \right) = \left ( \frac{3}{2} - \frac{3}{3} \right ) + \left ( \frac{3}{3} - \frac{3}{4} \right ) + \left ( \frac{3}{4} - \frac{3}{5} \right ) + ... + \left ( \frac{3}{n+1} - \frac{3}{n+2} \right ) \\ s_n = \frac{3}{2} - \frac{3}{n+2} \end{align}

Therefore the sequence of partial sums $\{ s_n \} = \{ \frac{3}{2} - \frac{3}{n+2} \}$. Taking the limit of this sequence we have that:

(11)
\begin{align} \quad \lim_{n \to \infty} s_n = \lim_{n \to \infty} \left ( \frac{3}{2} - \frac{3}{n+2} \right) = \frac{3}{2} \end{align}

Therefore $\sum_{n=1}^{\infty} \frac{3}{n^2 + 3n + 2} = \frac{3}{2}$.