What is the sum from n=1 to infinity of (1+2/n)^2e^{-5n} ?

The sum from n=1 to infinity of (1+2/n)^2e^{-5n} is converges

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sum from n=1 to infinity of (1+2/n)^2e^{-5n}

n = 1 \sum \infty (1 + \frac{2}{n})^{2} e^{- 5 n}

converges

Solution steps

n = 1 \sum \infty (1 + \frac{2}{n})^{2} e^{- 5 n}

Apply Series Ratio Test:

converges

= converges

5 x^{2} - 1 + 7 e^{x}

\frac{\partial}{\partial x} (\frac{3 y}{x})

y^{^{''}} + 6 y^{^{'}} + 25 = 0

y^{^{''}} + 6 y^{^{'}} + 10 y = 5 t - 2

\int_{1}^{3} π (\frac{1}{x} + 1)^{2} d x

What is the sum from n=1 to infinity of (1+2/n)^2e^{-5n} ?
The sum from n=1 to infinity of (1+2/n)^2e^{-5n} is converges