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

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 $\frac{1}{\sqrt e}$

$$e^x=\sum_{r=0}^\infty\dfrac{x^r}{r!}$$
• @Did, For $$\displaystyle\lim_{n\to\infty}\sum_{k=0}^n \left(-\frac12\right)^k=\frac 23$$ in the question – lab bhattacharjee Dec 7 '18 at 9:40
• @Did, I added that formula so that OP can compare that with that of $e^x$ – lab bhattacharjee Dec 7 '18 at 9:50
We have $\displaystyle e^x=\sum_{k=0}^\infty\frac{x^k}{k!}$.
So $\displaystyle\lim_{n\to\infty}\sum_{k=0}^n \frac1{k!}\left(-\frac12\right)^k=\displaystyle\lim_{n\to\infty}\sum_{k=0}^n \frac{\left(-\frac12\right)^k}{k!}=e^{\frac {-1}{2}}=\frac 1{\sqrt e}$