What is the sum from n=1 to infinity of 1/(n^{1.01)} ?

The sum from n=1 to infinity of 1/(n^{1.01)} 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 1/(n^{1.01)}

n = 1 \sum \infty \frac{1}{n ^{1.01}}

converges

Solution steps

n = 1 \sum \infty \frac{1}{n ^{1.01}}

Apply p-Series Test:

converges

= converges

\int 4 π y (6 + y^{\frac{3}{2}}) d y

y = 7 x^{2} + 3

\frac{d}{d x} (e^{x} sin (2 x) + cos (2 x) \cdot 2 e^{x})

x \to 3 lim (25)

\frac{d}{d x} ((x^{2} - 2 x + 1) (x^{3} - 1))

What is the sum from n=1 to infinity of 1/(n^{1.01)} ?
The sum from n=1 to infinity of 1/(n^{1.01)} is converges