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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply p-Series Test:

diverges

= - diverges

= diverges

\int e^{3 x^{2}} d x

\frac{\partial}{\partial θ} (r sin (θ))

\int \frac{2}{x ^{3} + x ^{2} + x + 1} d x

\int 7 x^{3} e^{5 x} d x

(1 + x)^{2}

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