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

The sum from n=1 to infinity of 5/(2n^2+1) 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 5/(2n^2+1)

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

= 5 \cdot n = 1 \sum \infty \frac{1}{2 n ^{2} + 1}

Apply Series Comparison Test:

converges

= 5 converges

= converges

\int \frac{1}{1} (2 x) d x

\frac{\partial}{\partial x} (x^{2} - 2 y^{2} + 4)

\int 4 x^{3} e^{x} d x

\frac{1}{x}, 0, y = 1, y = e

x \to 3 lim (x^{3} + 6 x)

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