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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 2 \sum \infty \frac{e ^{n}}{π ^{n}}

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

n = 0 \sum \infty (0.3)^{n}

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

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

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