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Simplify problem

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\[\sum_{n=1, n=odd}^{\infty} (i/(\pi*n))(-(i) ^{n}+(i) ^{-n})\]
the final answer contains something with (-1)^n in the numerator
  • 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
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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