What is the sum from n=1 to infinity of (n+1)/2 ?

The sum from n=1 to infinity of (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 (n+1)/2

n = 1 \sum \infty \frac{n + 1}{2}

diverges

Solution steps

n = 1 \sum \infty \frac{n + 1}{2}

Apply the constant multiplication rule:

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

= \frac{1}{2} \cdot n = 1 \sum \infty n + 1

Apply Series Divergence Test:

diverges

= \frac{1}{2} diverges

= diverges

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

k = 0 \sum \infty (\frac{1}{10})^{k}

k = 1 \sum \infty 3 (- \frac{1}{5})^{3 k}

n = 0 \sum \infty - \frac{3}{( n ^{2} - 4 )}

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

What is the sum from n=1 to infinity of (n+1)/2 ?
The sum from n=1 to infinity of (n+1)/2 is diverges