## anonymous 4 years ago Simplify problem

1. anonymous

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

2. anonymous

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

3. phi

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 $\frac{2}{i^n}$

4. phi

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 =$$\frac{1}{2}$$