Proving Sum of Fractions with a=1: A Study Guide

[quasar987]

Hi. I am starting the study of series and I don't see how to do this problem.

"Show that

$$\sum_{n=0}^\infty \frac{1}{(n+a)(n+1+a)} = \frac{1}{a}$$"

All i got is the decomposition in partial fractions as

$$\sum_{n=0}^\infty (...) = \sum_{n=0}^\infty \frac{1}{(n+a)} + \sum_{n=0}^\infty \frac{-1}{(n+1+a)}$$

if these sum converge. I tried seeing a patern in the partial sums to find $S_n$ but it's too difficult so there must be another way.


Any hint/help will be appreciated.

 

[e(ho0n3]

Using your partial fractions decomposition and rearranging the terms gives:

$$\frac{1}{a} - \frac{1}{a+1} + \frac{1}{a+1} - \frac{1}{a+2} + \ldots$$

It can't be any more obvious now can it.

 

[wais]

Hi there! It's great that you're starting to study series, it can definitely be a challenging topic. Let me try to help you out with this problem.

First, let's rewrite the given series as:

\sum_{n=0}^\infty \frac{1}{(n+a)(n+1+a)} = \sum_{n=0}^\infty \frac{1}{n+a} - \frac{1}{n+1+a}

Now, let's focus on the first term, \frac{1}{n+a}. We can rewrite this as \frac{1}{n+a} = \frac{1}{a} \cdot \frac{1}{n+1}, using the fact that a=1.

Substituting this into our series, we get:

\sum_{n=0}^\infty \frac{1}{(n+a)(n+1+a)} = \sum_{n=0}^\infty \frac{1}{a} \cdot \frac{1}{n+1} - \frac{1}{n+1+a}

Now, we can combine these two terms by finding a common denominator:

\frac{1}{a} \cdot \frac{1}{n+1} - \frac{1}{n+1+a} = \frac{1}{a(n+1)(n+1+a)} - \frac{a}{a(n+1)(n+1+a)}

Notice that the denominators are the same, so we can simply add the numerators:

\frac{1}{a(n+1)(n+1+a)} - \frac{a}{a(n+1)(n+1+a)} = \frac{1-a}{a(n+1)(n+1+a)}

Now, we can rewrite this as:

\frac{1}{a(n+1)(n+1+a)} = \frac{1}{a} \cdot \frac{1}{(n+a)(n+1+a)}

And this is exactly the first term in our original series! So, we can rewrite the series as:

\sum_{n=0}^\infty \frac{1}{(n+a)(n+1+a)} = \frac{1}{a} \sum_{n=0}^\infty \frac{1}{(n+a)(n+1+a)}

But this is just