What is the sum from n=0 to infinity}(i^{2n of)/(3n) ?

The sum from n=0 to infinity}(i^{2n of)/(3n) is converges

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sum from n=0 to infinity}(i^{2n of)/(3n)

n = 0 \sum \infty \frac{i ^{2 n}}{3 n}

converges

Solution steps

i = 1 \sum \infty \frac{n ^{2 i}}{3 i}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= \frac{1}{3} \cdot n = 1 \sum \infty \frac{i ^{2 n}}{n}

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

= \frac{1}{3} converges

= converges

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

n = 1 \sum \infty \frac{4}{n ^{2} + 2 n}

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

n = 0 \sum \infty (\frac{1}{2})^{2}

k = 1 \sum \infty \frac{1}{k ^{2} + k}

What is the sum from n=0 to infinity}(i^{2n of)/(3n) ?
The sum from n=0 to infinity}(i^{2n of)/(3n) is converges