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

The sum from n=1 to infinity of (-1)^nn^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 (-1)^nn^2

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

\int \frac{5 x ^{2} - 9}{x ^{3} - 9 x} d x

f (x) = \frac{x - 5}{x + 5}

f (x) = \frac{( 1 - 5 x )}{1 + 5 x}

\frac{d}{d x} y + y = 6

\int (2 - 2 x)^{9} d x

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