2
$\begingroup$

How to evaluate the limit $$\lim_{n \to +\infty}(-1)^n\frac{n^n + \ln n}{\cos(n\pi)(n + \pi)^n}$$?

I was suspecting that the sequence does not converge, but the ratio test, which is the only one that I know, turned out to be useless. The limit of the ratio $\frac{a_{n+1}}{a_n}$ is $1$, so the test is inconclusive.

It turns out the result is $\boxed{\displaystyle e^{-\pi}}$, but I fail to see how one would reach it.

Any suggestions?

$\endgroup$
  • 2
    $\begingroup$ $(-1)^n$ and $\cos(n\pi)$ cancel out, $\log n$ gives a smallish contribute. Then, what is the limit $$\lim_{n\to +\infty}\left(\frac{n}{\pi +n}\right)^n $$? $\endgroup$ – Jack D'Aurizio Jan 10 '15 at 11:47
  • $\begingroup$ @JackD'Aurizio: Thanks, I hadn't realized that $(-1)^n$ and $\cos(n\pi)$ represent the same thing! $\endgroup$ – rubik Jan 10 '15 at 12:03
6
$\begingroup$

Observe that $\cos(\pi n)=(-1)^n$, hence $$ (-1)^n\frac{n^n+\log n}{\cos(\pi n)(n+\pi)^n}=\frac{n^n+\log n}{(n+\pi)^n}=\frac{1+\frac{\log n}{n^n}}{\left(1+\frac{\pi}{n}\right)^n} $$ Now the numerator goes to $1$, and the denominator is a notable limit, it goes to $e^{\pi}$, hence you get what you wanted.

$\endgroup$
  • $\begingroup$ Great, thank you! Very clear. $\endgroup$ – rubik Jan 10 '15 at 11:59

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.