What is the sum from n=1 to infinity of (8-5)/(n^2) ?

The sum from n=1 to infinity of (8-5)/(n^2) is converges

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 (8-5)/(n^2)

n = 1 \sum \infty \frac{8 - 5}{n ^{2}}

converges

Solution steps

n = 1 \sum \infty \frac{8 - 5}{n ^{2}}

Apply the constant multiplication rule:

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

= (8 - 5) \cdot n = 1 \sum \infty \frac{1}{n ^{2}}

Simplify

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

Apply p-Series Test:

converges

= 3 converges

= converges

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

n = 1 \sum \infty \frac{5}{17 + 8 ^{n}}

n = 2 \sum \infty \frac{1}{6} (\frac{5}{6})^{n}

k = 1 \sum \infty \frac{5}{( k + 4 ) ^{2}}

n = 0 \sum \infty n^{\frac{1}{3}}

What is the sum from n=1 to infinity of (8-5)/(n^2) ?
The sum from n=1 to infinity of (8-5)/(n^2) is converges