# Validation of partial sum $\sum _{i=1}^{i=n}\dfrac {x} {\left( \left( i-1\right) x+1\right) \left( ix+1\right) }=1-\dfrac {1} {nx+1}$

I came across the below partial sum where the author did not provide any reasoning and claimed that it is easy to see that the partial sum is so $$\sum _{i=1}^{i=n}\dfrac {x} {\left( \left( i-1\right) x+1\right) \left( ix+1\right) }=1-\dfrac {1} {nx+1}$$ I want to validate this, although i am willing to bet that his claim must be right but, i could n't quite accomplish this. The link does infact confirm it.

Obviously the series converges, my original idea was to get a common denominator and sum up the terms in the numerators, but this ends up being a very laborious process.

Second thought was to try think of a well known partial sum or series and obtain the sum expression based on that, but i can not seem to match this to a pattern of a well known series as it stands atleast.

There seems to be a lot of tools which help with convergence but probably not many when it comes to exact computation. I was hoping if some one could please shed some light on how to validate this and also if we have any other tools in our arsenal to evaluate partial sums.

Any help would be much appreciated.

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Hint: telescope sum –  Blah Mar 15 '12 at 19:36
$$\frac{x}{((i-1)x+1)(ix+1)} = \frac1{(i-1)x+1)} - \frac1{ix+1}$$ –  user17762 Mar 15 '12 at 19:38
Thank you guys that was very helpfull. –  Hardy Mar 15 '12 at 19:43
@Hardy: Note that the expression that was used is just the partial fractions version of your expression. Partial fractions are not only for integration! Computer Science students end up having to know about partial fractions, because of their usefulness in dealing with common generating functions. –  André Nicolas Mar 15 '12 at 20:11
@Hardy: Yes, one gets combinatorial problems. Generating functions are often rational functions, and partial fractions often let us get explicit formulas for the coefficients. –  André Nicolas Mar 15 '12 at 22:12

Hint: $$\dfrac {x} {\left( \left( i-1\right) x+1\right) \left( ix+1\right) }={\dfrac {1} {\left( \left( i-1\right) x+1\right)} } - {\dfrac {1} {\left(i) x+1\right)} }$$