What is the sum from n=1 to infinity of n/(3^{n^2)} ?

The sum from n=1 to infinity of n/(3^{n^2)} is converges

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sum from n=1 to infinity of n/(3^{n^2)}

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

f (x) = 5 x^{2} e^{3 x}

y = (1 + 2 x)^{2}, (3, 49)

\frac{s}{( s + 1 ) ^{2} + 4}

\frac{\partial}{\partial y} (x^{2} + y^{2} + z^{2})

\int 5 cos^{3} (2 x) d x

What is the sum from n=1 to infinity of n/(3^{n^2)} ?
The sum from n=1 to infinity of n/(3^{n^2)} is converges