anonymous
  • anonymous
Simplify problem
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\sum_{n=1, n=odd}^{\infty} (i/(\pi*n))(-(i) ^{n}+(i) ^{-n})\]
anonymous
  • anonymous
the final answer contains something with (-1)^n in the numerator
phi
  • 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 \[ \frac{2}{i^n} \]

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phi
  • 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 =\( \frac{1}{2} \)

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