Today I've been studying series convergence and I'm blocked by a problem as follow :

It's a question about the convergence of $a_n$. They give $a_n= \sum_{k=0}^n \frac{2^k}{k!} $ if n is even and $ a_n = (\sum_{k=0}^n \frac{1}{k!})^c$ if n is odd. Then they ask what we can tell about it. The answer needed is that there is only one value to c that allow $a_n$ to converge but it's hard for me to come to a definitive answer. My guess is that since both are convergent, the whole thing converge and I can more or less understand that c can impact it but it's hard for me to visualize "mathematically"

For the "even part' I clearly see that it's convergent, but for the odd part it's harder :

I know that $\sum_{k=0}^n \frac{1}{k!}$ converge but I wonder how I could calculate with the c.

And a minor remark but since we are asked about the whole $a_n$, does both part need to converge to the same value ? I think so right ?

Thanks for your help !

  • $\begingroup$ From the definition of "the sequence $a_n$ is convergent" you can see that if $a_n\to\ell$ then both $a_{2n}\to\ell$ and $a_{2n+1}\to\ell$; so if you get different limits from the odd and even terms the sequence itself doesn't converge. So you are right in the last paragraph. $\endgroup$ Jan 9, 2022 at 15:02
  • $\begingroup$ Hint: do you know the series for $e^x$? If so, what is $\lim a_{2n}$? do you know that if $b_n\to\ell$ then $b_n^c\to\ell^c$? $\endgroup$ Jan 9, 2022 at 15:04
  • $\begingroup$ Oh wow I've never thought about using the taylor series thanks a lot ! $\endgroup$ Jan 9, 2022 at 15:38

1 Answer 1


The Taylor/Maclaurin series of $\ e^x\ $ is:

$$f(x)=e^x=\sum _{k=0}^{\infty}\frac{x^k}{k!},$$

and this holds true for all $\ x.$

Therefore, $\ f(2) = e^2 = \displaystyle\sum _{k=0}^{\infty}\frac{2^k}{k!}.$

Therefore we have $\ \displaystyle\lim_{n\ \text{is even};\ n\to\infty} a_{n} = e^2,\ $ so in order for $\ a_n\ $ to converge, $\ a_n\ $ must converge to $\ e^2,\ $ and so we must also have $\ \displaystyle\lim_{n\ \text{is odd};\ n\to\infty} a_{n} = e^2,\ \implies \lim_{n\to\infty}\left(\displaystyle\sum _{k=0}^{n}\frac{1}{k!}\right)^c = e^c = e^2 \implies c=2.$

  • $\begingroup$ Yeah that's it, it never crossed my mind that it could be the MacLaurin series of some sort, it make much more senses now, thank you very much $\endgroup$ Jan 9, 2022 at 15:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .