What is the sum from n=6 to infinity of 1/((n-5)^2) ?

The sum from n=6 to infinity of 1/((n-5)^2) is converges

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sum from n=6 to infinity of 1/((n-5)^2)

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

converges

Solution steps

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

Apply Series Integral Test:

converges

= converges

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

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

n = 1 \sum \infty \frac{π}{n ^{2}}

n = 1 \sum \infty (\frac{- 1}{3})^{n}

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

What is the sum from n=6 to infinity of 1/((n-5)^2) ?
The sum from n=6 to infinity of 1/((n-5)^2) is converges