What is the sum from n=0 to infinity}(i^{2n of)/n ?

The sum from n=0 to infinity}(i^{2n of)/n is converges

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sum from n=0 to infinity}(i^{2n of)/n

n = 0 \sum \infty \frac{i ^{2 n}}{n}

converges

Solution steps

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

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

L > 1,

by the Raabe's test

= converges

n = 2 \sum \infty (- 0.9)^{n}

n = 1 \sum \infty \frac{n !}{2 n}

n = 1 \sum \infty n^{- 0.6}

n = 0 \sum \infty \frac{4}{9 - n ^{2}}

k = 1 \sum \infty \frac{3}{k} - \frac{3}{k + 2}

What is the sum from n=0 to infinity}(i^{2n of)/n ?
The sum from n=0 to infinity}(i^{2n of)/n is converges