What is the sum from n=1 to infinity of ((4-3)^n)/n ?

The sum from n=1 to infinity of ((4-3)^n)/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 ((4-3)^n)/n

n = 1 \sum \infty \frac{( 4 - 3 ) ^{n}}{n}

diverges

Solution steps

n = 1 \sum \infty \frac{( 4 - 3 ) ^{n}}{n}

Simplify

= n = 1 \sum \infty \frac{1}{n}

Apply Cauchy's Convergence Condition:

diverges

= diverges

n = 1 \sum \infty \frac{8 ^{n}}{n ^{k}}

n = 1 \sum \infty sin^{n} (8)

n = 1 \sum \infty \frac{1}{5 ^{n} + 3}

n = 1 \sum \infty \frac{n !}{10 4 ^{n}}

n = 1 \sum \infty \frac{2}{n ( n + 4 )}

What is the sum from n=1 to infinity of ((4-3)^n)/n ?
The sum from n=1 to infinity of ((4-3)^n)/n is diverges