1
$\begingroup$

So I was trying to find the values of $p>0$ such that the sum $$\sum_{n=1}^{\infty} \left(1-\frac{p}{n}\right)^n $$

converges. I tried to use the root test but the limit of that equals $1$ so the root test is inconclusive. How else would I do this question though? I can't use the integral test or the limit comparison test. The ratio test also fails.

$\endgroup$
  • $\begingroup$ From Bernoulli's Inequality, $\left(1-\frac pn\right)^n\ge 1-p$. $\endgroup$ – Mark Viola Feb 27 '18 at 1:07
1
$\begingroup$

It fails to converge as the $n$th terms don't tend to $0$: $$\lim_{n\to\infty}\left(1-\frac{p}{n}\right)^n = e^{-p}$$

$\endgroup$
  • $\begingroup$ Yes of course. I can't believe I missed that. Thank you! $\endgroup$ – Future Math person Feb 27 '18 at 1:07
2
$\begingroup$

Hint: what is

$$ \lim_{n \to \infty} \left( 1 - \frac{p}{n} \right)^n ? $$

$\endgroup$
  • $\begingroup$ Yes I completely forgot about the definition of e. Thanks! $\endgroup$ – Future Math person Feb 27 '18 at 1:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.