What is the sum from n=1 to infinity of 3/(n^{1/2)} ?

The sum from n=1 to infinity of 3/(n^{1/2)} 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 3/(n^{1/2)}

n = 1 \sum \infty \frac{3}{n ^{\frac{1}{2}}}

diverges

Solution steps

n = 1 \sum \infty \frac{3}{n ^{\frac{1}{2}}}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty \frac{1}{n ^{\frac{1}{2}}}

Apply p-Series Test:

diverges

= 3 diverges

= diverges

n = 0 \sum \infty \frac{2}{n ^{3} + 5}

n = 0 \sum \infty cos (\frac{1}{n + 1})

k = 6 \sum \infty \frac{1}{( k - 5 ) ^{4}}

n = 1 \sum \infty 53 \cdot (\frac{1}{2})^{n}

n = 0 \sum \infty \frac{1}{ln ( 6 n )}

What is the sum from n=1 to infinity of 3/(n^{1/2)} ?
The sum from n=1 to infinity of 3/(n^{1/2)} is diverges