What is the sum from n=1 to infinity of 5/(4^n+1) ?

The sum from n=1 to infinity of 5/(4^n+1) is converges

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/(4^n+1)

n = 1 \sum \infty \frac{5}{4 ^{n} + 1}

converges

Solution steps

n = 1 \sum \infty \frac{5}{4 ^{n} + 1}

Apply the constant multiplication rule:

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

= 5 \cdot n = 1 \sum \infty \frac{1}{4 ^{n} + 1}

Apply Series Ratio Test:

converges

= 5 converges

= converges

n = 0 \sum \infty \frac{1}{6 ^{n} - 5}

n = 3 \sum \infty \frac{1}{n ( n - 2 )}

n = 5 \sum \infty \frac{1}{n ^{2} ln ( n )}

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

n = 1 \sum \infty \frac{7}{n ^{\frac{2}{5}}}

What is the sum from n=1 to infinity of 5/(4^n+1) ?
The sum from n=1 to infinity of 5/(4^n+1) is converges