Calculate sum of $\sum_{n=3} \frac{1}{4(n-2)}-\frac{1}{4(n+2)}$

Question

I have the following telescopic series:

$$\sum_{n=3} \frac{1}{4(n-2)}-\frac{1}{4(n+2)}$$

I want to calculate its sum. I'm assuming this is a telescopic series of the following type:

$$\sum_{n=1}{a_n-a_{a+k}}$$

So the sum should theoretically be $a_1 + a_2 + \dots + a_k$ with $k = 4$.

In this case, though, $n=3$, so how do I calculate the sum of this series? Any hints?

Answer 1

You can change index by replacing $k=n-2$ to get

$$\sum_{k=1}^{\infty} \frac 1{4k} - \frac 1{4(k+4)}$$

Is this more familiar now?

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For more rigorous proof I would like you to refer to partial sums. Then as we have cancelation after the $4$ term we get:

$$S_N = \sum_{k=1}^{N} \frac 1{4k} - \frac 1{4(k+4)} = \frac 14 - \frac 1{20} + \frac 18 - \frac 1{24} + \frac 1{12} - \frac 1{28} + \frac 1{16} - \frac 1{32} + \frac{1}{20} - \frac{1}{36} + ... $$

$$+\frac 1{4N} - \frac 1{4(N+4)}= \frac 14 + \frac 18 + \frac 1{12} + \frac 1{16} - \frac 1{4(N+1)} - \frac 1{4(N+2)} - \frac 1{4(N+3)} - \frac{1}{4(N+3)}$$

Now we have that:

$$\sum_{k=1}^{\infty} \frac 1{4k} - \frac 1{4(k+4)} = \lim_{N \to \infty} S_N = \frac 14 + \frac 18 + \frac 1{12} + \frac 1{16} = \frac{25}{48}$$

as the last four terms in $S_N$ tend to $0$, as $N \to \infty$

Answer 2

The sum is $ = \frac{1}{4}\left[1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right]$

Expand

$\frac{1}{1}\ - \frac{1}{5}$

$\frac{1}{2}\ - \frac{1}{6}$

$\frac{1}{3}\ - \frac{1}{7}$

$\frac{1}{4}\ - \frac{1}{8}$

$\frac{1}{5}\ - \frac{1}{9}$

...

The $\frac{1}{5}$ in the first term cancels out with the $\frac{1}{5}$ in the fifth term and hence forth all terms in the right gets cancelled to yeild the above. $$ = \frac{25}{48}$$