What is the sum from n=0 to infinity of 7/(n+3) ?

The sum from n=0 to infinity of 7/(n+3) 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 7/(n+3)

n = 0 \sum \infty \frac{7}{n + 3}

diverges

Solution steps

n = 0 \sum \infty \frac{7}{n + 3}

Apply the constant multiplication rule:

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

= 7 \cdot n = 0 \sum \infty \frac{1}{n + 3}

Apply Series Integral Test:

diverges

= 7 diverges

= diverges

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

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

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

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

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

What is the sum from n=0 to infinity of 7/(n+3) ?
The sum from n=0 to infinity of 7/(n+3) is diverges