What is the sum from n=1 to infinity of (5/3)^n ?

The sum from n=1 to infinity of (5/3)^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 (5/3)^n

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

diverges

Solution steps

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

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= n = 0 \sum \infty (\frac{5}{3})^{n} - (\frac{5}{3})^{0}

n = 0 \sum \infty (\frac{5}{3})^{n} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

n = 1 \sum \infty (\frac{7 π}{8})^{n}

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

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

n = 0 \sum \infty \frac{4}{n ^{2} + 1}

n = 0 \sum \infty \frac{n ^{2}}{4 ^{n}}

What is the sum from n=1 to infinity of (5/3)^n ?
The sum from n=1 to infinity of (5/3)^n is diverges