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Popular >

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

Solution

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

Solution

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

Popular Examples

y^{^{'}} - \frac{1}{x} y = 1

tangent of y= 1/(4+3x),(1, 1/7)tangent of

y = \frac{1}{4 + 3 x}, (1, \frac{1}{7})

f^'(x)=sqrt(2-4)

f^{^{'}} (x) = 2 - 4

integral of \sqrt[3]{x^6}

derivative of e^xsec(x)

\frac{d}{d x} (e^{x} sec (x))

Frequently Asked Questions (FAQ)

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