Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Find $$\lim_{n \to \infty}\left[ {1 \over x + 1} + {2x \over \left(x + 1\right)\left(x + 2\right)} +{3x^{2} \over \left(x + 1\right)\left(x + 2\right)\left(x + 3\right)} + \cdots + {nx^{n-1} \over \left(x + 1\right)\left(x + 2\right)\ldots\left(x + n\right)}\right]$$

share|cite|improve this question
it should be x^{n-1} – DeepSea May 12 '14 at 3:07
Can partial fractions help in any way? – rah4927 May 12 '14 at 3:30
up vote 4 down vote accepted


For integer $r\ge1$,


$$\iff \frac{rx^{r-1}}{(x+1)(x+2)\cdots(x+r-1)(x+r)}$$ $$=\frac{x^{r-1}}{(x+1)(x+2)\cdots(x+r-1)}-\frac{x^r}{(x+1)(x+2)\cdots(x+r-1)(x+r)} $$

Set $r=1,2,3,c\dots, n-1,n$ to add to recognize the Telescoping Series

Finally set $n\to\infty$

share|cite|improve this answer
Thanks. Good one :) – Uma kant May 12 '14 at 4:11
@Umakant, Welcome. But, what is limit? – lab bhattacharjee May 12 '14 at 4:18
It comes down to $\lim_{n\rightarrow \infty} 1 - \frac{x^n}{(x+1)(x+2)...(x+n)} = 1$. Let me know if I am wrong. – Uma kant May 12 '14 at 4:25
@Umakant, It's true for finite $x$ if $x$ is $O(n^r)$ where $r<1$ – lab bhattacharjee May 12 '14 at 12:49
Even if my $x\rightarrow \infty$ and $n\rightarrow \infty$, I guess the above limit will give 1. Or is there some thing else? – Uma kant May 13 '14 at 8:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.