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

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

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

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

converges

Solution steps

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

Apply Series Integral Test:

converges

= converges

n = 0 \sum \infty 5 (- \frac{3}{2})^{n}

n = 15 \sum \infty (\frac{2}{3})^{n}

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

n = 1 \sum \infty 12 sin^{n} (7 0^{\circ})

n = 0 \sum \infty \frac{n}{5 n ^{2} + 1}

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