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

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

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

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

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

Solution steps

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

Apply the constant multiplication rule:

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

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

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

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

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

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

Simplify

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

n = 1 \sum \infty \frac{1}{( 7 n + 4 ) ^{2}}

n = 0 \sum \infty n^{5}

n = 1 \sum \infty 3 n^{- 2}

n = 0 \sum \infty \frac{n ^{4}}{n + 2}

n = 2 \sum \infty \frac{n - 1}{n !}

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