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Evaluate:$\displaystyle\lim_{n\to\infty}\sum_{k=0}^n \frac1{k!}\left(-\frac12\right)^k$

MY TRY:We know that $\displaystyle\lim_{n\to\infty}\sum_{k=0}^n \frac1{k!}=e$ and $\displaystyle\lim_{n\to\infty}\sum_{k=0}^n \left(-\frac12\right)^k=\frac 23$ but how can we evaluate the above$?$Thank you.

Note:The answer is $\frac{1}{\sqrt e}$

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HINT:

$$e^x=\sum_{r=0}^\infty\dfrac{x^r}{r!}$$

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  • $\begingroup$ Why the second hint? $\endgroup$ – Did Dec 7 '18 at 9:35
  • $\begingroup$ @Did, For $$\displaystyle\lim_{n\to\infty}\sum_{k=0}^n \left(-\frac12\right)^k=\frac 23$$ in the question $\endgroup$ – lab bhattacharjee Dec 7 '18 at 9:40
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    $\begingroup$ No, this is mentioned as a try and we know it leads nowhere. Adding it to your answer can only mislead the OP to think they need this fact to solve their question, while they do not. $\endgroup$ – Did Dec 7 '18 at 9:45
  • $\begingroup$ @Did, I added that formula so that OP can compare that with that of $e^x$ $\endgroup$ – lab bhattacharjee Dec 7 '18 at 9:50
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We have $\displaystyle e^x=\sum_{k=0}^\infty\frac{x^k}{k!}$.

So $\displaystyle\lim_{n\to\infty}\sum_{k=0}^n \frac1{k!}\left(-\frac12\right)^k=\displaystyle\lim_{n\to\infty}\sum_{k=0}^n \frac{\left(-\frac12\right)^k}{k!}=e^{\frac {-1}{2}}=\frac 1{\sqrt e}$

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