What is the sum from n=0 to infinity of (e/2)^n ?

The sum from n=0 to infinity of (e/2)^n 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=0 to infinity of (e/2)^n

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

diverges

Solution steps

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

Apply Series Geometric Test:

diverges

= diverges

\int x ln (x + 3) d x

\int_{1}^{e} 8 t^{2} ln (t) d t

f (x) = x^{2} + 6 x - 1

y^{^{'''}} - y^{^{''}} - y^{^{'}} + y = 5

\int 7 cot^{5} (x) sin^{4} (x) d x

What is the sum from n=0 to infinity of (e/2)^n ?
The sum from n=0 to infinity of (e/2)^n is diverges