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Sum of $$\frac{1}{(x+1)}+\frac{2x}{(x+1)(x+2)}+\frac{3x^2}{(x+1)(x+2)(x+3)}+\ldots$$ till n terms ...

I have no clue how to solve this ... I tried multiplying quantities ... such as $(x+2)-(x+1)$

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Have you tried partial fractions? – Stephen Herschkorn Aug 5 '13 at 0:08
up vote 6 down vote accepted

We have $$ \frac{kx^{k-1}}{(x+1)(x+2)\cdots(x+k-1)(x+k)}\\= \frac{x^{k-1}}{(x+1)(x+2)\cdots(x+k-1)}- \frac{x^{k}}{(x+1)(x+2)\cdots(x+k)},\, k\geq2$$ so by telescoping $$\sum_{k=1}^{n} \frac{kx^{k-1}}{(x+1)(x+2)\cdots(x+k-1)(x+k)}=\frac{1}{x+1}+\frac{x}{x+1}-\frac{x^{n}}{(x+1)(x+2)\cdots(x+n-1)(x+n)}\\=1-\frac{x^{n}}{(x+1)(x+2)\cdots(x+n-1)(x+n)}$$

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Of course, the first two terms on the RHS sum to 1. – Stephen Herschkorn Aug 5 '13 at 0:35
Yes. Thanks for the comment. – user63181 Aug 5 '13 at 0:37

Note that,

$$1-\prod_{k=0}^n\frac{1}{a_k+1}=\sum_{k=0}^na_k\prod_{j=0}^k\frac{1}{a_j+1}$$ For all $\{a_i \}$ with $a_i\ne -1$ and integers $n\ge 0$,

Which can easily be verified by mathematical induction

Now if you set $a_k=\frac{k}{x}$, and simplify the denominators the result readily follows.

Also as an interesting side note the first identity is used to derive some beautiful $q$-series identities, and was first used by Euler to derive the pentagonal number theorem.

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