Question

Simplify problem

Answer 1

\[\sum_{n=1, n=odd}^{\infty} (i/(\pi*n))(-(i) ^{n}+(i) ^{-n})\]

Answer 2

the final answer contains something with (-1)^n in the numerator

Answer 3

We could first simplify the expression in paretheses
\[ - i^n + i^{-n} \]
find a common denominator, and add:
\[\frac{1}{i^n}-\frac{i^{2n}}{i^n}= \frac{1- (i^2)^n}{i^n}= \frac{1- (-1)^n}{i^n}\]
n is odd so we know -1^(odd_power) is always going to be -1. Therefore the expression becomes
\[ \frac{2}{i^n} \]

Answer 4

using the above in your summation\[\sum_{n=1, n=odd}^{\infty} (i/(\pi*n))(-(i) ^{n}+(i) ^{-n})= \frac{2}{\pi}\sum_{n=1, odd}^{\infty}\frac{1}{n i^{n-1}}\]
I do not like summation indices that go up by 2, so replace the summation index n with m=0,1,2,...
n= 2m+1
re-write the summation to get
\[\frac{2}{\pi}\sum_{m=0}^{\infty}\frac{1}{(2m+1) i^{2m}}=\frac{2}{\pi}\sum_{m=0}^{\infty}\frac{1}{(2m+1) (i^{2})^{m}}\]
of course, i^2 is -1, so this becomes a series with alternating signs:
\[\frac{2}{\pi}(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}...)\]
This is a well-known series
http://mathworld.wolfram.com/LeibnizSeries.html
You can easily find the final answer =\( \frac{1}{2} \)