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.

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}

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