What is the sum from n=0 to infinity of n/((n+1)^2) ?

The sum from n=0 to infinity of n/((n+1)^2) is diverges

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sum from n=0 to infinity of n/((n+1)^2)

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

diverges

Solution steps

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

(n + 1)^{2} = n^{2} + 2 n + 1

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

Apply Series Integral Test:

diverges

= diverges

n = 0 \sum \infty \frac{n + 1}{3 n + 1}

n = 0 \sum \infty e^{- 14 n}

n = 1 \sum \infty \frac{n}{2 n ^{2} + 11}

n = 1 \sum \infty \frac{6}{9 n + 1}

n = 1 \sum \infty \frac{5}{n ^{2} + 4 n + 3}

What is the sum from n=0 to infinity of n/((n+1)^2) ?
The sum from n=0 to infinity of n/((n+1)^2) is diverges