Series help!; converge or diverge

**badBKO** · November 23rd 2008, 04:12 PM

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

Does this converge or diverge and how?

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

I'm leaning on converge for this one because

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

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

**Mathstud28** · November 23rd 2008, 05:08 PM

Originally Posted by badBKO

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

Does this converge or diverge and how?

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

I'm leaning on converge for this one because

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

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

$\sum_{n=1}^{\infty}\frac{n^n}{(\pi-2)^{4n^2}}$

What about the Root test?

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

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

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

Right idea! consider

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

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

$\begin{aligned}\lim_{z\to{0}}\frac{\sinh(z)}{z}&=\ lim_{z\to{0}}\frac{\sin(iz)}{iz}\\<br /> &=1\end{aligned}$

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

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

**badBKO** · November 23rd 2008, 05:19 PM

Thanks alot!!!!

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

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