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How to find the limit $\displaystyle\lim_{n \to + \infty}{S_n}$ where $$S_n= \dfrac{1}{2} + \dfrac{1}{2\cdot 4} + \dfrac{1}{2\cdot 4 \cdot 6} + \dfrac{1}{2\cdot 4 \cdot 6\cdots 2n}? $$ I know that $\dfrac{1}{2}<S_n<1. $ With Maple, I guess the answer $.6487212707$, but I can not explain.

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You can get a hint from Maple: > identify(.6487212707); $$-1 + {\rm e}^{1/2}$$ –  Robert Israel Jan 6 '13 at 9:04
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1 Answer 1

up vote 6 down vote accepted

$$t_r=\frac1{2.4.6.2r}=\frac1 {r!}\left(\frac12\right)^r$$

$$S_n=\sum_{r=1}^nt_r=\sum_{r=1}^n\frac1 {r!}\left(\frac12\right)^r$$

So, $\lim_{n\to \infty}S_n=\sum_{r=1}^ \infty\frac1 {r!}(\frac12)^r=\sum_{r=0}^ \infty\frac1 {r!}(\frac12)^r-1=\text{e}^{\frac12}-1$

Google says, $$\sqrt{\text{e}}=1.6487212707$$

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