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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

n = 1 \sum \infty \frac{1}{n ^{1.43}}

n = 1 \sum \infty \frac{9}{7 n ^{2} + 7 n}

n = 0 \sum \infty \frac{1}{n 2 ^{n}}

n = 1 \sum \infty 9 ln (\frac{n}{n + 1})

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

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