What is the sum from k=2 to infinity of 1/((k-1)^5) ?

The sum from k=2 to infinity of 1/((k-1)^5) is converges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from k=2 to infinity of 1/((k-1)^5)

k = 2 \sum \infty \frac{1}{( k - 1 ) ^{5}}

converges

Solution steps

k = 2 \sum \infty \frac{1}{( k - 1 ) ^{5}}

Apply Series Integral Test:

converges

= converges

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

n = 0 \sum \infty \frac{9}{10}

n = 2 \sum \infty 4 n^{- 2}

n = 0 \sum \infty (\frac{1}{13})^{n}

n = 6 \sum \infty \frac{1}{( n - 5 ) ^{2}}

What is the sum from k=2 to infinity of 1/((k-1)^5) ?
The sum from k=2 to infinity of 1/((k-1)^5) is converges