What is the sum from n=1 to infinity of n/(4^n) ?

The sum from n=1 to infinity of n/(4^n) is converges

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sum from n=1 to infinity of n/(4^n)

n = 1 \sum \infty \frac{n}{4 ^{n}}

converges

Solution steps

n = 1 \sum \infty \frac{n}{4 ^{n}}

Apply Series Ratio Test:

converges

= converges

\frac{d}{d x} (6 x - 4)

y^{^{''}} + y^{^{'}} + y = x e^{x} + e^{- x} (1 + 2 x)

\int \frac{e ^{- x}}{1 + e ^{- 2 x}} d x

f (x) = 8 sin^{4} (x)

\frac{d}{d x} (6 x^{5} - 15 x^{4} + 10 x^{3})

What is the sum from n=1 to infinity of n/(4^n) ?
The sum from n=1 to infinity of n/(4^n) is converges