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

The sum from n=2 to infinity of 1/(2n*ln(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=2 to infinity of 1/(2n*ln(n))

n = 2 \sum \infty \frac{1}{2 n \cdot ln ( n )}

diverges

Solution steps

n = 2 \sum \infty \frac{1}{2 n ln ( n )}

Apply the constant multiplication rule:

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

= \frac{1}{2} \cdot n = 2 \sum \infty \frac{1}{n ln ( n )}

Apply Series Integral Test:

diverges

= \frac{1}{2} diverges

= diverges

x^{- 2} (5 x^{2} - 4)^{3}

(2 x^{3} + x y^{2}) + (x^{2} y + 2 y^{3}) \frac{d y}{d x} = 0

\int_{0}^{t} e^{- s^{2}} d s

x \to - 5 lim (\frac{x ^{2} - 2}{5 - x})

\int \frac{x ^{3}}{3} \cdot \frac{1}{x} d x

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