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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

\frac{d}{d x} (2 x^{- 3})

f (x) = \frac{4}{x}

\int \frac{3 x ^{4} - x ^{3} + 6 x ^{2}}{x ^{4}} d x

y = 12 - x^{2}, y = x^{2} - 6

\frac{d}{d x} ((cos (7 x))^{x})

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