What is the sum from n=0 to infinity of (3^n)/(4^n) ?

The sum from n=0 to infinity of (3^n)/(4^n) is converges

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sum from n=0 to infinity of (3^n)/(4^n)

n = 0 \sum \infty \frac{3 ^{n}}{4 ^{n}}

converges

Solution steps

n = 0 \sum \infty \frac{3 ^{n}}{4 ^{n}}

n \to \infty lim (n (\frac{\frac{3 ^{n}}{4 ^{n}}}{\frac{3 ^{n + 1}}{4 ^{n + 1}}} - 1)) = \infty

L > 1,

by the Raabe's test

= converges

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

\int_{1}^{\infty} (\frac{1}{x ^{2}}) d x

x \to \infty lim (f (x) (x)^{2})

x \to - \infty lim (x^{2} - 1)

(x^{2} + 1) (x^{2} - 2 x)

What is the sum from n=0 to infinity of (3^n)/(4^n) ?
The sum from n=0 to infinity of (3^n)/(4^n) is converges