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

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

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 (e^n)/pi

n = 1 \sum \infty \frac{e ^{n}}{π}

diverges

Solution steps

n = 1 \sum \infty \frac{e ^{n}}{π}

Apply the constant multiplication rule:

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

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

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

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

n = 0 \sum \infty \frac{1}{n ^{5} + 1}

k = 1 \sum \infty \frac{1}{k ^{5}}

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

n = 0 \sum \infty \frac{6}{n}

n = 1 \sum \infty n (\frac{5}{6})^{n}

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