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

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

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

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

converges

Solution steps

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

Apply Series Comparison Test:

converges

= converges

n = 0 \sum \infty 3 (\frac{1}{3})^{n}

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

x = 1 \sum \infty \frac{ln ( x )}{x}

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

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

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