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

Question

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

Possible duplicate of How does $e^0 = 1$ if you define $e^x = \sum_{n = 0}^\infty x^n/n!$ — Guy Fsone, Dec 29 '17 at 19:26
https://math.stackexchange.com/questions/441836/how-do-we-know-that-sum-k-0-infty-fracxkk-ex — Guy Fsone, Dec 29 '17 at 19:28

lab bhattacharjee · Accepted Answer · 2018-12-07T09:49:13.847

3

HINT:

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

edited Dec 07 '18 at 09:49

answered Jan 14 '17 at 12:58

lab bhattacharjee

274,582

Why the second hint? – Did Dec 07 '18 at 09:35
@Did, For $$\displaystyle\lim_{n\to\infty}\sum_{k=0}^n \left(-\frac12\right)^k=\frac 23$$ in the question – lab bhattacharjee Dec 07 '18 at 09:40
1

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. – Did Dec 07 '18 at 09:45
@Did, I added that formula so that OP can compare that with that of $e^x$ – lab bhattacharjee Dec 07 '18 at 09:50

score 1 · Answer 2 · answered Jan 14 '17 at 13:17

We have $\displaystyle e^x=\sum_{k=0}^\infty\frac{x^k}{k!}$.

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

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

2 Answers2