MHB 242.tr.05 Use the integral test to determine if a series converges.

[karush]

$\tiny{242.tr.05}$
Use the integral test to determine
if a series converges.
$\displaystyle
\sum_{n=1}^{\infty}\frac{1}{\sqrt{e^{2n}-1}}$
so...
$\displaystyle
\int_{1}^{\infty} \frac{1}{\sqrt{e^{2n}-1}}\, dn
=\int_{1}^{\infty} (e^{2n}-1)^{1/2} \, dn $
so
$u=e^{2n}-1\therefore du=2e^{2n}$

 

[MarkFL]

I would define:

$$S=\sum_{n=1}^{\infty}\frac{1}{\sqrt{e^{2n}-1}}$$

And so:

$$S<\int_0^{\infty}\frac{1}{\sqrt{e^{2x}-1}}\,dx$$

Let:

$$u=e^{x}\implies dx=\frac{du}{u}$$

And so:

$$S<\int_1^{\infty}\frac{1}{u\sqrt{u^2-1}}\,du$$

Let:

$$u=\sec(\theta)\implies du=\sec(\theta)\tan(\theta)\,d\theta$$

And so:

$$S<\int_0^{\frac{\pi}{2}}\,d\theta=\frac{\pi}{2}$$

Thus, the series converges.

edit: I played fast and loose with some improper integrals...:p

 

[Greg]

$$\int_0^\infty\frac{1}{\sqrt{e^{2x}-1}}\,dx$$

$$e^{2x}-1=z^2$$

$$2e^{2x}\,dx=2z\,dz$$

$$dx=\frac{z}{z^2+1}\,dz$$

$$\int_0^\infty\frac{1}{z^2+1}\,dz=\lim_{z\to\infty}\arctan(z)=\frac{\pi}{2}$$