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

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

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

diverges

Solution steps

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

Factor

2 - n : - (- 2 + n)

= n = 3 \sum \infty \frac{1}{- ( - 2 + n )}

Apply the constant multiplication rule:

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

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

Apply Series Comparison Test:

diverges

= - diverges

= diverges

n = 0 \sum \infty (\frac{5}{7})^{n}

n = 1 \sum \infty \frac{n !}{9 ^{n}}

k = 1 \sum \infty \frac{4}{k} - \frac{4}{k + 1}

n = 1 \sum \infty \frac{1}{3 n ^{2} + 8}

k = 6 \sum \infty \frac{1}{k ^{2} - 5 k + 6}

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