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

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it should be x^{n-1} –  OC-Sansoo May 12 at 3:07
1  
Can partial fractions help in any way? –  rah4927 May 12 at 3:30

1 Answer 1

up vote 4 down vote accepted

HINT:

For integer $r\ge1$,

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

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

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

Finally set $n\to\infty$

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Thanks. Good one :) –  Uma kant May 12 at 4:11
    
@Umakant, Welcome. But, what is limit? –  lab bhattacharjee May 12 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 at 4:25
    
@Umakant, It's true for finite $x$ if $x$ is $O(n^r)$ where $r<1$ –  lab bhattacharjee May 12 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 at 8:17

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