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

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

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

n = 1 \sum \infty (\frac{e}{π})^{n}

converges

Solution steps

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

y = x^{3}, (1, 1)

\frac{d}{d t} (- \frac{1}{t + 1})

\frac{d}{d x} ((\frac{x}{x + 1})^{5})

t \to 0 lim (\frac{sin ( k t )}{t})

f (x) = \frac{x}{a}

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