What is the sum from n=0 to infinity}(i^{2n of)/(4n) ?

The sum from n=0 to infinity}(i^{2n of)/(4n) is converges

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sum from n=0 to infinity}(i^{2n of)/(4n)

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

= \frac{1}{4} \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}{4} converges

= converges

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

n = 1 \sum \infty \frac{( 3 n - 2 )}{n}

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

n = 1 \sum \infty \frac{4 ^{n}}{n ^{7}}

k = 1 \sum \infty \frac{1}{k ^{7}}

What is the sum from n=0 to infinity}(i^{2n of)/(4n) ?
The sum from n=0 to infinity}(i^{2n of)/(4n) is converges