View Full Version : Infinite series
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