What is the sum from n=5 to infinity of (n+1)/(n^3) ?

The sum from n=5 to infinity of (n+1)/(n^3) is converges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

sum from n=5 to infinity of (n+1)/(n^3)

n = 5 \sum \infty \frac{n + 1}{n ^{3}}

converges

Solution steps

n = 5 \sum \infty \frac{n + 1}{n ^{3}}

Expand

\frac{n + 1}{n ^{3}} : \frac{1}{n ^{2}} + \frac{1}{n ^{3}}

= n = 5 \sum \infty \frac{1}{n ^{2}} + \frac{1}{n ^{3}}

Apply the Sum Rule:

\sum a_{n} + b_{n} = \sum a_{n} + \sum b_{n}

= n = 5 \sum \infty \frac{1}{n ^{2}} + n = 5 \sum \infty \frac{1}{n ^{3}}

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

converges

n = 5 \sum \infty \frac{1}{n ^{3}} =

converges

= converges + converges

= converges

What is the sum from n=5 to infinity of (n+1)/(n^3) ?
The sum from n=5 to infinity of (n+1)/(n^3) is converges