Telescoping Series Examples 2

Telescoping Series Examples 2

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 1 page.

Example 1

The series $\sum_{n=1}^{\infty} \frac{1}{3^n} - \frac{1}{3^{n+1}}$ converges. Find its sum.

We note that the $n^{\mathrm{th}}$ partial sum $s_n$ is given by:

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

Therefore $\lim_{n \to \infty} s_n = \frac{1}{3}$ and so $\sum_{n=1}^{\infty} \frac{1}{3^n} - \frac{1}{3^{n+1}} = \frac{1}{3}$.

Note that we could have alternatively computed the sum with geometric series, however, the sum would have required more work to compute.

Example 2

The series $\sum_{n=1}^{\infty} \frac{3}{n(n+2)}$ converges. Find its sum.

Applying the techniques of partial fractions and we have that:

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

Therefore $3 = 3A(n+2) + 3B(n)$ or rather $1 = A(n+2) + B(n)$. If $n = 0$ then we see that $A = \frac{1}{2}$. If $n = -2$ then we see that $B = \frac{1}{2}$, and so:

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

Now the $n^{\mathrm{th}}$ partial sum can be computed as:

(4)
\begin{align} \quad \quad s_n = \left ( \frac{3}{2} + \frac{3}{4} + \frac{3}{6} + ... + \frac{3}{2n} \right ) - \left ( \frac{3}{6} + \frac{3}{8} + ... + \frac{3}{2n} + \frac{3}{2n + 2} + \frac{3}{2n + 4} \right ) \\ \quad \quad s_n = \frac{3}{2} + \frac{3}{4} - \frac{3}{2n+2} - \frac{3}{2n+4} \end{align}

Therefore $\lim_{n \to \infty} s_n = \frac{3}{2} + \frac{3}{4} = \frac{9}{4}$ and so $\sum_{n=1}^{\infty} \frac{3}{n(n+2)} = \frac{9}{4}$.