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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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}

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

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

= \frac{e}{e - 1} - \frac{1}{e ^{0}}

Simplify

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

\int \frac{x ^{2} - 8}{8 x} d x

\frac{d}{d x} (ln (arctan (9 x^{4})))

f (x) = ln (x)

x \to 0 lim (x ln (3 x))

k \to 0 lim (\frac{( 2 z + 3 k ) ^{3} - 4 k ^{2} z}{2 z ( 2 z - k ) ^{2}})

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