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

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

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

diverges

Solution steps

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

Apply Series Integral Test:

diverges

= diverges

\frac{d}{d x} (3 x^{\frac{1}{4}})

\frac{d}{d x} (2 x^{3} - 3 x^{2} - 72 x + 3)

y = ln (sec (x) + tan (x))

\frac{d}{d x} ((2 x - 45) (x^{2} + 162))

\frac{d}{d x} (sin^{3} (cos^{2} (e^{x^{2} + 1})))

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