What is the sum from n=0 to infinity of-4/(e^n) ?

The sum from n=0 to infinity of-4/(e^n) is -4*e/(e-1)

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sum from n=0 to infinity of-4/(e^n)

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

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

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Geometric Test:

\frac{e}{e - 1}

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

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

n = 1 \sum \infty \frac{n + 3}{2 n + 1}

k = 0 \sum \infty (- \frac{1}{5})^{k}

n = 1 \sum \infty \frac{8 - 5}{n ^{2}}

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

What is the sum from n=0 to infinity of-4/(e^n) ?
The sum from n=0 to infinity of-4/(e^n) is -4*e/(e-1)