What is the sum from n=0 to infinity of (5^n)/(n^2) ?

The sum from n=0 to infinity of (5^n)/(n^2) is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from n=0 to infinity of (5^n)/(n^2)

n = 0 \sum \infty \frac{5 ^{n}}{n ^{2}}

diverges

Solution steps

n = 1 \sum \infty \frac{5 ^{n}}{n ^{2}}

n \to \infty lim (n (\frac{\frac{5 ^{n}}{n ^{2}}}{\frac{5 ^{n + 1}}{( n + 1 ) ^{2}}} - 1)) = - \infty

L < 1,

by the Raabe's test

= diverges

n = 0 \sum \infty \frac{8}{( 1 - 7 n )}

n = 1 \sum \infty 53 (\frac{1}{2})^{n}

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

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

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

What is the sum from n=0 to infinity of (5^n)/(n^2) ?
The sum from n=0 to infinity of (5^n)/(n^2) is diverges