What is the sum from n=0 to infinity of (11)/(2n-1) ?

The sum from n=0 to infinity of (11)/(2n-1) 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=0 to infinity of (11)/(2n-1)

n = 0 \sum \infty \frac{11}{2 n - 1}

diverges

Solution steps

n = 0 \sum \infty \frac{11}{2 n - 1}

Apply the constant multiplication rule:

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

= 11 \cdot n = 0 \sum \infty \frac{1}{2 n - 1}

Apply Series Integral Test:

diverges

= 11 diverges

= diverges

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

n = 1 \sum \infty \frac{1}{2} + \frac{4}{3 ^{n}}

n = 1 \sum \infty 3 n (\frac{1}{4})^{n}

n = 0 \sum \infty cos (2 nπ)

n = 0 \sum \infty \frac{15}{4} (\frac{1}{4})^{n}

What is the sum from n=0 to infinity of (11)/(2n-1) ?
The sum from n=0 to infinity of (11)/(2n-1) is diverges