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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

= \frac{1}{2} \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}{2} converges

= converges

k = 1 \sum \infty \frac{1}{k k}

n = 0 \sum \infty 7 + \frac{5}{n ^{5}}

n = 1 \sum \infty (1.05)^{n}

n = 1 \sum \infty \frac{n}{4 n ^{3} + 3}

k = 2 \sum \infty (\frac{3}{5})^{2 k}

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