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

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

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 i/(n^2)

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply p-Series Test:

converges

= i converges

= converges

k = 1 \sum \infty (cos (14))^{k}

n = 1 \sum \infty \frac{3}{n} - \frac{3}{n + 1}

n = 0 \sum \infty n e^{- 2 n^{2}}

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

n = 0 \sum \infty 9 (\frac{8}{9})^{n}

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