What is the sum from n=1 to infinity of 2/(e^{n-3)} ?

The sum from n=1 to infinity of 2/(e^{n-3)} is 2e^3(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 2/(e^{n-3)}

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

2 e^{3} (\frac{e}{e - 1} - 1)

Solution steps

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

Apply the constant multiplication rule:

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

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

Simplify

e^{n - 3} : e^{n} e^{- 3}

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

Apply the constant multiplication rule:

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

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

\frac{1}{e ^{- 3}} = e^{3}

= 2 e^{3} \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}

= 2 e^{3} (n = 0 \sum \infty \frac{1}{e ^{n}} - \frac{1}{e ^{0}})

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

= 2 e^{3} (\frac{e}{e - 1} - \frac{1}{e ^{0}})

Simplify

= 2 e^{3} (\frac{e}{e - 1} - 1)

y^{^{''}} - \frac{2}{x} y^{^{'}} - \frac{4}{x ^{2}} y = 3 x^{2}

x \to 1 + lim (x^{3} - 4)

x \to - 1 lim (\frac{2 x}{x + 1})

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

\frac{d}{d x} (\frac{1}{arctan ( x )})

What is the sum from n=1 to infinity of 2/(e^{n-3)} ?
The sum from n=1 to infinity of 2/(e^{n-3)} is 2e^3(e/(e-1)-1)