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

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

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

diverges

Solution steps

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

Simplify

\frac{n}{n} : n

= n = 1 \sum \infty n

Apply Series Divergence Test:

diverges

= diverges

f (x) = 4^{x} x^{4}

\frac{e ^{- 2 s}}{s - 2}

\int \frac{1}{( x ^{2} 4 - x ^{2} )} d x

\int y^{\frac{1}{2}} d y

\int e^{100 t} \cdot \frac{sin ( 100 t )}{0.01} d t

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