Q 64. - Calculus

Question

Show that the series $\sum _{k=1}^{\infty }\frac{1}{k\left(k+1\right)\left(k+2\right)}$ is a telescoping series. Also, provide the general term $S_n$ in its sequence of partial sums and find the sum of the series if it converges.

Step-by-Step Solution

Answer

The series $\sum _{k=1}^{\infty }\frac{1}{k\left(k+1\right)\left(k+2\right)}$ has general term $S_n=\frac{1}{4}-\frac{1}{2n\left(n+1\right)}$ and sum of the series is $\frac{1}{4}$.

Step 1. Given information.

Given a series $\sum _{k=1}^{\infty }\frac{1}{k\left(k+1\right)\left(k+2\right)}$.

Step 2. Find if the series is telescoping series or not.

The kth term of this series has the partial-fraction decomposition $\frac{1}{k\left(k+1\right)\left(k+2\right)}=\frac{1}{2k\left(k+1\right)}-\frac{1}{2\left(k+1\right)\left(k+2\right)}$.

Consider $a_k=\frac{1}{2k\left(k+1\right)}$.

It follows that $\sum _{k=1}^{\infty }\frac{1}{k\left(k+1\right)\left(k+2\right)}=\sum _{k=1}^{\infty }\left(a_k-a_{k+1}\right)$.

So, $S_n=\sum _{n=1}^{\infty }\left(a_n-a_{n+1}\right)$.

Note that between each two consecutive pairs, the second term of a pair cancels with the first term of the subsequent pair. 

The cancellation between these values gives rise to the term “telescoping” series in that each term in the sequence of partial sums collapses like a

folding telescope.

Now, $S_n=a_1-a_n$, that is $S_n=\frac{1}{4}-\frac{1}{2n\left(n+1\right)}$.

Therefore, the series is a telscoping series.

Step 3. Find the sum of the series if it converges.

As n tends to infinity, $a\to 0$.

It follows that $S_n\to a_1$.

Note that $a_1=\frac{1}{4}$.

Therefore, sum of the series is $\frac{1}{4}$.