Find if the series converges or diverges.

Question

anonymous · Answer

$$\sum_{n=1}^{\infty} \frac{ (2n+1)^{n} }{ n ^{2n} }$$

anonymous · Answer

I've tried ratio test and root test, didn't have success with either. Don't know where to go form here.

anonymous · Answer

You'll have better luck with the root test.

anonymous · Answer

I couldn't get it. i got $$\frac{ 2n+1 }{ n^2 }$$ which is just inf/inf

anonymous · Answer

Actually...that limit equals zero doesn't it? Because the degree on the bottom is larger than the top.

anonymous · Answer

Yes the limit is zero.

anonymous · Answer

Yep, forgot Calc 1 stuff somehow. Thanks, I appreciate it.

anonymous · Answer

yw

mathmath333 · Answer

i thought u hve to prove the limit

anonymous · Answer

We just have to check for conv/divergence of the series. We have tests for that. No limit proofs needed, unless we want to make absolutely sure that $\dfrac{2n+1}{n^2}\to0$.

anonymous · Answer

Yep, it converges as 0 < 1 due to the root test conditions of convergence

mathmath333 · Answer

$\large\tt \begin{align} \color{black}{\dfrac{2n+1}{n^2}\
=\dfrac{2}{n}+\dfrac{1}{n^2}\
=\dfrac{2}{\infty}+\dfrac{1}{\infty^2}\
=0+0=0}\end{align}$