What is the sum from n=3 to infinity of 1/(esqrt(n)) ?

The sum from n=3 to infinity of 1/(esqrt(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=3 to infinity of 1/(esqrt(n))

n = 3 \sum \infty \frac{1}{e n}

diverges

Solution steps

n = 3 \sum \infty \frac{1}{e n}

Apply the constant multiplication rule:

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

= \frac{1}{e} \cdot n = 3 \sum \infty \frac{1}{n}

Apply p-Series Test:

diverges

= \frac{1}{e} diverges

= diverges

\frac{d y}{d x} = 0.0013 y (2500 - y)

\frac{d y}{d t} = 100 - y

\int_{0}^{3} 9

\int x (15 x + 8) d x

\frac{\partial}{\partial x} (\frac{3}{x y ^{2}})

What is the sum from n=3 to infinity of 1/(esqrt(n)) ?
The sum from n=3 to infinity of 1/(esqrt(n)) is diverges