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

The sum from k=5 to infinity of (e/pi)^{k-1} is converges

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sum from k=5 to infinity of (e/pi)^{k-1}

k = 5 \sum \infty (\frac{e}{π})^{k - 1}

converges

Solution steps

k = 5 \sum \infty (\frac{e}{π})^{k - 1}

Apply Series Ratio Test:

converges

= converges

k = 4 \sum \infty \frac{1}{( k - 1 ) ^{4}}

n = 0 \sum \infty 6 (\frac{2}{3})^{n}

n = 0 \sum \infty 2 (\frac{1}{2})^{n}

n = 1 \sum \infty (\frac{3}{4})^{n - 1}

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

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