# Thread: Series help!; converge or diverge

1. ## Series help!; converge or diverge

$\displaystyle \sum_{n=1} ^\infty (\sqrt[3]{n})^{3n} * {(\Pi-2)}^{-4n^2}$

Does this converge or diverge and how?

$\displaystyle \sum_{n=1} ^\infty {sinh(1/e^x)}$

I'm leaning on converge for this one because

the limit of $\displaystyle (1/e^x)$ = 0
so,

$\displaystyle \sum_{n=1} ^\infty {sinh(1/e^x)}$
Converges by the Limit Comparison Test?

$\displaystyle \sum_{n=1} ^\infty (\sqrt[3]{n})^{3n} * {(\Pi-2)}^{-4n^2}$

Does this converge or diverge and how?

$\displaystyle \sum_{n=1} ^\infty {sinh(1/e^x)}$

I'm leaning on converge for this one because

the limit of $\displaystyle (1/e^x)$ = 0
so,

$\displaystyle \sum_{n=1} ^\infty {sinh(1/e^x)}$
Converges by the Limit Comparison Test?
$\displaystyle \sum_{n=1}^{\infty}\frac{n^n}{(\pi-2)^{4n^2}}$

\displaystyle \begin{aligned}\lim_{n\to\infty}\sqrt[n]{\frac{n^n}{(\pi-2)^{4n^2}}}&=\lim_{n\to\infty}\frac{n}{(\pi-2)^{4n}}\\ &=0<1\end{aligned}

The last part was gotten since the exponential function increases faster than any polynomial. So this series is convergent

$\displaystyle \sum_{n=0}^{\infty}\sinh\left(e^{-x}\right)$

Right idea! consider

$\displaystyle \lim_{x\to\infty}\frac{\sinh\left(e^{-x}\right)}{e^{-x}}$

Making the sub $\displaystyle z=e^{-x}$ gives

\displaystyle \begin{aligned}\lim_{z\to{0}}\frac{\sinh(z)}{z}&=\ lim_{z\to{0}}\frac{\sin(iz)}{iz}\\ &=1\end{aligned}

The last part can be made with the sub $\displaystyle iz=\varphi$

So both $\displaystyle \sum_{n=0}^{\infty}\frac{1}{e^x}$ and $\displaystyle \sum_{n=0}^{\infty}\sinh\left(\frac{1}{e^x}\right)$ share convergence.

3. Thanks alot!!!!

The first one didn't look that easy, the second one, I thought I was on the right track.