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$$\displaystyle \lim_{n \to \infty} \sum_{i=1}^n \frac{n+2}{2(n-1)!}$$

Is there a theorem to be applied here? Or is there way to use telescoping series?

Please help.

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2 Answers 2

up vote 7 down vote accepted

Assuming the problem to be $\displaystyle \lim_{n\to\infty}\sum_{1\le i\le n}\frac{i+2}{2\cdot (i-1)!}$

$$\text{Now, }\frac{r+2}{2\cdot(r-1!)}=\frac{r-1+3}{2\cdot(r-1!)}=\frac12\frac1{(r-2)!}+\frac32\frac1{(r-1)!}$$

We know from here, $\displaystyle e=\sum_{0\le r<\infty}\frac1{r!}$.

In general using Partial Fraction Decomposition, we can write

$$\displaystyle\frac{a_0+a_1r+a_2r^2+\cdots+a_mr^m}{(r-u)!}$$ $$=\frac{b_0+b_1(r-u)+b_2(r-u)(r-u-1)+\cdots+b_m(r-u)(r-u-1)\cdots\{r-u-(m-1)\}}{(r-u)!}$$


Multiply either by $(r-u)!$ to get

$$a_0+a_1r+a_2r^2+\cdots+a_mr^m=b_0+b_1(r-u)+b_2(r-u)(r-u-1)+\cdots$$ $$+b_m(r-u)(r-u-1)\cdots\{r-u-(m-1)\}$$

Compare the constants and the coefficients of the different powers of $r$ to find $b_i$s

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Which theorem states that? – jay Nov 22 '13 at 8:47

$$\sum_{n=1}^\infty{n+2\over 2(n-1)!}=\sum_{n=0}^\infty{n+3\over 2n!}={1\over2}\sum_{n=1}^\infty{1\over (n-1)!}+{3\over2}\sum_{n=0}^\infty{1\over n!}=2e\ .$$

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