What is the sum from n=1 to infinity of (3^n)/(n^2) ?

The sum from n=1 to infinity of (3^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=1 to infinity of (3^n)/(n^2)

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

diverges

Solution steps

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

n \to \infty lim (n (\frac{\frac{3 ^{n}}{n ^{2}}}{\frac{3 ^{n + 1}}{( n + 1 ) ^{2}}} - 1)) = - \infty

L < 1,

by the Raabe's test

= diverges

\int - sec^{2} (x) d x

\int \frac{t ^{2} + 1}{t} d t

\int \frac{3}{x ^{\frac{1}{2}}} d x

n = 0 \sum \infty 2

\int_{0}^{2} \frac{8 t}{( t - 3 ) ^{2}} d t

What is the sum from n=1 to infinity of (3^n)/(n^2) ?
The sum from n=1 to infinity of (3^n)/(n^2) is diverges