1
$\begingroup$

The problem

Hi, I’ve been trying to crack this problem but I’m unable to make any progress, could you please give me hints? Thx

Edit: The solution is from wolframalpha

$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

Simply use

$$ \sum_{i=0}^{\infty} \frac{(an)^ie^{-an}e^{-i/n}}{i!}=e^{-an} \sum_{i=0}^{\infty} \frac{(ane^{-1/n})^i}{i!}=e^{-an} e^{ane^{-1/n}} $$

$\endgroup$
1
  • $\begingroup$ Thank you so much, I should revisit stuff like this as I’ve apparently forgotten a lot. $\endgroup$
    – innerz09
    Commented May 18, 2019 at 9:46

You must log in to answer this question.

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