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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

(x + 2)^{2} + (y - 3)^{2} = 37, (- 1, - 3)

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

y = \frac{x + x ^{3}}{x ^{2}}

f (x) = 7 - 2 x

\frac{\partial}{\partial x} (2 x^{3} + 4 y - 2 z^{2})

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