What is the sum from k=2 to infinity of 1/((k-1)^4) ?

The sum from k=2 to infinity of 1/((k-1)^4) is converges

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sum from k=2 to infinity of 1/((k-1)^4)

k = 2 \sum \infty \frac{1}{( k - 1 ) ^{4}}

converges

Solution steps

k = 2 \sum \infty \frac{1}{( k - 1 ) ^{4}}

Apply Series Integral Test:

converges

= converges

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

n = 2 \sum \infty \frac{1}{n ^{\frac{5}{3}}}

n = 2 \sum \infty (\frac{2}{7})^{n}

n = 0 \sum \infty n (n - 6)

n = 1 \sum \infty 6 (\frac{1}{8})^{n - 1}

What is the sum from k=2 to infinity of 1/((k-1)^4) ?
The sum from k=2 to infinity of 1/((k-1)^4) is converges