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

The sum from n=3 to infinity of (n^2)/(e^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=3 to infinity of (n^2)/(e^n)

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

0.03 x^{2} - 4 x + 3 x^{0.8} - 5000

\int - \frac{t}{1 + t ^{2}} d t

f (x) = 14 cos (x), at x = 7

y^{^{'}} + ln (x) = \frac{y}{x}

\int_{0}^{1} 1 + (- \frac{1}{x})^{2} d x

What is the sum from n=3 to infinity of (n^2)/(e^n) ?
The sum from n=3 to infinity of (n^2)/(e^n) is converges