What is the sum from n=0 to infinity of (n+1)/(e^n) ?

The sum from n=0 to infinity of (n+1)/(e^n) is converges

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sum from n=0 to infinity of (n+1)/(e^n)

n = 0 \sum \infty \frac{n + 1}{e ^{n}}

converges

Solution steps

n = 0 \sum \infty \frac{n + 1}{e ^{n}}

Expand

\frac{n + 1}{e ^{n}} : \frac{n}{e ^{n}} + \frac{1}{e ^{n}}

= n = 0 \sum \infty \frac{n}{e ^{n}} + \frac{1}{e ^{n}}

Apply the Sum Rule:

\sum a_{n} + b_{n} = \sum a_{n} + \sum b_{n}

= n = 0 \sum \infty \frac{n}{e ^{n}} + n = 0 \sum \infty \frac{1}{e ^{n}}

n = 0 \sum \infty \frac{n}{e ^{n}} =

converges

n = 0 \sum \infty \frac{1}{e ^{n}} = \frac{e}{e - 1}

= converges + \frac{e}{e - 1}

= converges

\int (1 - x^{2})^{2} d x

3 6 x^{4} + 7 x^{3}

\frac{d}{d x} ((6 x + 4) (3 x^{2} + 2))

x \to - \infty lim (\frac{5 x ^{4} + e ^{5 x}}{x ^{4}})

\frac{d ^{2}}{d x ^{2}} ((5 + \frac{2}{x})^{3})

What is the sum from n=0 to infinity of (n+1)/(e^n) ?
The sum from n=0 to infinity of (n+1)/(e^n) is converges