What is the sum from n=5 to infinity of 1/((n-2)^4) ?

The sum from n=5 to infinity of 1/((n-2)^4) is converges

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sum from n=5 to infinity of 1/((n-2)^4)

n = 5 \sum \infty \frac{1}{( n - 2 ) ^{4}}

converges

Solution steps

n = 5 \sum \infty \frac{1}{( n - 2 ) ^{4}}

Apply Series Integral Test:

converges

= converges

n = 2 \sum \infty 9 e^{- 2 n}

n = 0 \sum \infty e^{2 l n (n)}

k = 1 \sum \infty \frac{6}{k} - \frac{6}{k + 1}

n = 0 \sum \infty 5 (\frac{3}{π})^{n}

n = 1 \sum \infty (\frac{- 1}{7})^{n}

What is the sum from n=5 to infinity of 1/((n-2)^4) ?
The sum from n=5 to infinity of 1/((n-2)^4) is converges