What is the sum from n=1 to infinity of (100)/n ?

The sum from n=1 to infinity of (100)/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 (100)/n

n = 1 \sum \infty \frac{100}{n}

diverges

Solution steps

n = 1 \sum \infty \frac{100}{n}

Apply the constant multiplication rule:

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

= 100 \cdot n = 1 \sum \infty \frac{1}{n}

Apply Cauchy's Convergence Condition:

diverges

= 100 diverges

= diverges

\int e^{3 x - 2 y} + x^{2} e^{- 2 y} d x

\int \frac{( sin ( x ) + 1 ) ^{\frac{3}{2}}}{sec ( x )} d x

\int 1 + (\frac{3 x}{2})^{2} d x

f (x) = 6 x^{2} + 2 x - 6

\frac{d}{d x} (\frac{x + 1}{x - 3})

What is the sum from n=1 to infinity of (100)/n ?
The sum from n=1 to infinity of (100)/n is diverges