What is the sum from n=0 to infinity of (n+1)^2 ?

The sum from n=0 to infinity of (n+1)^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 (n+1)^2

n = 0 \sum \infty (n + 1)^{2}

diverges

Solution steps

n = 0 \sum \infty (n + 1)^{2}

Apply Series Divergence Test:

diverges

= diverges

\frac{\partial}{\partial x} (\frac{x ^{2} + y}{y ^{2}})

y^{^{''}} - 4 y^{^{'}} + 8 y = 0

f (x) = arcsin (\frac{x}{4})

\int x^{- \frac{1}{6}} d x

\int_{- 2}^{2} \frac{1}{4 - x ^{2}} d x

What is the sum from n=0 to infinity of (n+1)^2 ?
The sum from n=0 to infinity of (n+1)^2 is diverges