What is the sum from n=1 to infinity of n/(6n^2-5) ?

The sum from n=1 to infinity of n/(6n^2-5) 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=1 to infinity of n/(6n^2-5)

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

diverges

Solution steps

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

Apply Series Comparison Test:

diverges

= diverges

x \to + 4 - lim (\frac{2 x}{16 - x ^{2}})

y^{^{'}} = y^{2} (1 + x^{3})

\int (3 x + 2 e^{x})^{2} d x

\frac{\partial}{\partial x} (5 9 x^{2} + 16 y^{2})

\int \frac{7}{x ( x ^{2} + 25 )} d x

What is the sum from n=1 to infinity of n/(6n^2-5) ?
The sum from n=1 to infinity of n/(6n^2-5) is diverges