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}}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

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

Apply Series Ratio Test:

converges

= e converges

= converges

n = 1 \sum \infty ln (n^{2})

n = 1 \sum \infty n e^{2 n}

n = 1 \sum \infty \frac{( 1 )}{n}

n = 0 \sum \infty (\frac{n + 5}{n})^{n}

n = 1 \sum \infty \frac{1}{ln ( n ^{2} )}

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