What is the sum from n=1 to infinity of (e^n)/(4^n) ?

The sum from n=1 to infinity of (e^n)/(4^n) is 4/(4-e)-1

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sum from n=1 to infinity of (e^n)/(4^n)

n = 1 \sum \infty \frac{e ^{n}}{4 ^{n}}

\frac{4}{4 - e} - 1

Solution steps

n = 1 \sum \infty \frac{e ^{n}}{4 ^{n}}

Apply exponent rule:

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

= n = 1 \sum \infty (\frac{e}{4})^{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}{4})^{n} - (\frac{e}{4})^{0}

n = 0 \sum \infty (\frac{e}{4})^{n} = \frac{4}{4 - e}

= \frac{4}{4 - e} - (\frac{e}{4})^{0}

Simplify

= \frac{4}{4 - e} - 1

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

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

n = 1 \sum \infty cos (n \cdot π)

n = 5 \sum \infty \frac{1}{n}

n = 1 \sum \infty cos (n) π

What is the sum from n=1 to infinity of (e^n)/(4^n) ?
The sum from n=1 to infinity of (e^n)/(4^n) is 4/(4-e)-1