What is the sum from n=1 to infinity of (2^n)/((n)!) ?

The sum from n=1 to infinity of (2^n)/((n)!) is converges

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^n)/((n)!)

n = 1 \sum \infty \frac{2 ^{n}}{( n ) !}

converges

Solution steps

n = 1 \sum \infty \frac{2 ^{n}}{n !}

Apply Series Ratio Test:

converges

= converges

\frac{d}{d x} (x^{2} + 9 x)

\frac{\partial}{\partial x} (e^{x^{2} - y^{2}} (cos (2 x y) + sin (2 x y)))

4 x - 3 x^{2}

- 30 t^{2} + 380 t, 7, 14

\int (5 - 15 x^{2}) csc^{2} (x^{3} - x) d x

What is the sum from n=1 to infinity of (2^n)/((n)!) ?
The sum from n=1 to infinity of (2^n)/((n)!) is converges