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

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

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

n = 1 \sum \infty \frac{e ^{n}}{π ^{n}}

converges

Solution steps

n = 1 \sum \infty \frac{e ^{n}}{π ^{n}}

Apply exponent rule:

\frac{a ^{c}}{b ^{c}} = (\frac{a}{b})^{c}

= n = 1 \sum \infty (\frac{e}{π})^{n}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= n = 0 \sum \infty (\frac{e}{π})^{n} - (\frac{e}{π})^{0}

n = 0 \sum \infty (\frac{e}{π})^{n} =

converges

= converges - (\frac{e}{π})^{0}

= converges

\frac{d}{d x} (\frac{5 ^{x^{3}}}{4 x ^{2}})

y y^{^{'}} = \frac{7}{x}

(1 + e^{2 x}) y^{2} d y = e^{x} d x, y (0) = 0

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

f (x) = 5 - ∣ x - 5 ∣

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