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

The sum from k=n to infinity of (3/4)^k(1/4) is converges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Ratio Test:

converges

= \frac{1}{4} converges

= converges

n = 0 \sum \infty 4 (\frac{8}{9})^{n}

n = 1 \sum \infty cos (n)

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

n = 0 \sum \infty n^{2} - 2

n = 0 \sum \infty \frac{5 n}{10 + n}

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