What is the sum from n=1 to infinity of (23)/(2n) ?

The sum from n=1 to infinity of (23)/(2n) 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 (23)/(2n)

n = 1 \sum \infty \frac{23}{2 n}

diverges

Solution steps

n = 1 \sum \infty \frac{23}{2 n}

Apply the constant multiplication rule:

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

= \frac{23}{2} \cdot n = 1 \sum \infty \frac{1}{n}

Apply Cauchy's Convergence Condition:

diverges

= \frac{23}{2} diverges

= diverges

5 e^{(- 2 t)}

\int_{2}^{3} \frac{x}{x ^{2} - 3} d x

e^{a x + b}

\int sec^{2} (2 x) tan^{5} (2 x) d x

y = 4 x^{2} + 3 x

What is the sum from n=1 to infinity of (23)/(2n) ?
The sum from n=1 to infinity of (23)/(2n) is diverges