What is the sum from n=1 to infinity of (i^n)/(n^2) ?

The sum from n=1 to infinity of (i^n)/(n^2) is converges

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sum from n=1 to infinity of (i^n)/(n^2)

n = 1 \sum \infty \frac{i ^{n}}{n ^{2}}

converges

Solution steps

i = 1 \sum \infty \frac{n ^{i}}{i ^{2}}

n \to \infty lim (n (\frac{\frac{i ^{n}}{n ^{2}}}{\frac{i ^{n + 1}}{( n + 1 ) ^{2}}} - 1)) = \infty

L > 1,

by the Raabe's test

= converges

k = 1 \sum \infty \frac{1}{( k - 1 ) ^{3}}

n = 0 \sum \infty 1 2^{n}

n = 1 \sum \infty \frac{1}{1 ^{n}}

n = 1 \sum \infty \frac{n}{3 n ^{3} + 5}

n = 1 \sum \infty \frac{7}{n ( n + 7 )}

What is the sum from n=1 to infinity of (i^n)/(n^2) ?
The sum from n=1 to infinity of (i^n)/(n^2) is converges