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galactus · Jan 19, 2007, 11:25 AM

I've always liked infinite series. Here's one maybe you'll like to tackle. Not too bad.

Find the sum of:

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

It does converge.

galactus · Jan 21, 2007, 06:28 AM

It appears no one wants to take a stab.

Anyway:

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

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:

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

MathMaven53 · Feb 27, 2007, 10:44 AM

I thought you were summing (-1)^(n-1) (6n)/e^(2n) from n =1 to infinity
The problem you were solving is summing (-1)^(n-1) (6^n)/e^(2n)

My problem involves summing n r^n from n=1 to infinity for some number r
Hint: call this sum S. Then multiply both sides by r. Let m = n + 1 and replace n in result for rS. Write S as r plus sum of rest of terms with index of summation m. Compute rS - S
You have left a geometric series minus r. Sum geometric series and simplify right side.
Solve rS - S for S