What is the sum from n=1 to infinity of (n+1)/(n3^n) ?

The sum from n=1 to infinity of (n+1)/(n3^n) is converges

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sum from n=1 to infinity of (n+1)/(n3^n)

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

converges

Solution steps

n = 1 \sum \infty \frac{n + 1}{n \cdot 3 ^{n}}

Apply Series Ratio Test:

converges

= converges

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

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

k = 3 \sum \infty \frac{1}{( k - 2 ) ^{3}}

n = 0 \sum \infty \frac{i ^{2 n}}{2 n}

k = 1 \sum \infty \frac{1}{k k}

What is the sum from n=1 to infinity of (n+1)/(n3^n) ?
The sum from n=1 to infinity of (n+1)/(n3^n) is converges