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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

k = 6 \sum \infty \frac{1}{( k - 5 ) ^{4}}

n = 1 \sum \infty 53 \cdot (\frac{1}{2})^{n}

n = 0 \sum \infty \frac{1}{ln ( 6 n )}

n = 1 \sum \infty \frac{1}{12 n + 1}

k = 0 \sum \infty \frac{1}{k}

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