infinit series

**rcmango** · January 15th 2007, 10:51 PM

find the sum of the infinit series, if the it converges.

infinity ((-1)^n-1(6n)) / (e^2n)
E
n = 1

I'm confused because of the natural log in the denominator.
does it converge or diverge?

i just get errors while graphing.
How can i prove this mathematically.

thanks for the help.

**galactus** · January 16th 2007, 02:13 PM

Is this it:

$\sum_{n=1}^{\infty}\frac{(-1)^{n}-6n}{e^{2n}}$

If so, it converges.

Actually, it converges to:

$\frac{3}{2(sinh(1))^{2}}-\frac{1}{e^{2}+1}\approx{-1.20529541347}$

**rcmango** · January 16th 2007, 02:21 PM

Almost had it, just forget the n-1 part, instead of just n

here it is in image format:

there ya go.

**galactus** · January 16th 2007, 03:03 PM

Let's do it this way then:

$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}6^{n}}{e^{2n}}$

Let $S=\frac{6}{e^{2}}-\frac{6^{2}}{e^{4}}+\frac{6^{3}}{e^{6}}-\frac{6^{4}}{e^{8}}-..................$

Factor out $\frac{6}{e^{2}}$

$S=\frac{6}{e^{2}}\left(1-\underbrace{\frac{6}{e^{2}}+\frac{6^{2}}{e^{4}}-\frac{6^{3}}{e^{6}}-...............}_{\text{this equals S}}\right)$

Then we have:

$\frac{e^{2}}{6}S=1-S$

$\left(\frac{e^{2}}{6}+1\right)S=1$

Solve for S and we find S=

$S=\frac{6}{e^{2}+6}$

Therefore, it converges to:

$\frac{6}{e^{2}+6}\approx{0.448127183549}$

**Soroban** · January 16th 2007, 03:57 PM

Hello, rcmango!

Galactus did his usual excellent job.
. . We can make it a wee bit shorter . . .

Find the sum of the infinite series: . $\sum^{\infty}_{n=1}\frac{(\text{-}1)^{n-1}6^n}{e^{2n}}$

We have: . $\frac{6}{e^2} - \frac{6^2}{e^4} + \frac{6^3}{e^6} - \frac{6^4}{e^8} + \hdots$

This is a geometric series with first term: $a = \frac{6}{e^2}$ .and common ratio: $r = \text{-}\frac{6}{e^2}$

Therefore, its sum is: . $\frac{6}{e^2}\cdot\frac{1}{1 - \left(\text{-}\frac{6}{e^2}\right)} \;=\;\frac{6}{e^2+6}$

**rcmango** · January 16th 2007, 06:52 PM

Ya, thanks very much galactus!

You really broke it down nice so i could understand it fully.

appreciated this very much.

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