What is the sum from n=1 to infinity of ((-5)^n)/(3^{2n+1)} ?

The sum from n=1 to infinity of ((-5)^n)/(3^{2n+1)} is converges

Popular >

sum from n=1 to infinity of ((-5)^n)/(3^{2n+1)}

n = 1 \sum \infty \frac{( - 5 ) ^{n}}{3 ^{2 n + 1}}

converges

Solution steps

n = 1 \sum \infty \frac{( - 5 ) ^{n}}{3 ^{2 n + 1}}

Simplify

3^{2 n + 1} : 3^{2 n} \cdot 3

= n = 1 \sum \infty \frac{( - 5 ) ^{n}}{3 ^{2 n} \cdot 3}

Apply the constant multiplication rule:

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

= \frac{1}{3} \cdot n = 1 \sum \infty \frac{( - 5 ) ^{n}}{3 ^{2 n}}

3^{2 n} = 9^{n}

= \frac{1}{3} \cdot n = 1 \sum \infty \frac{( - 5 ) ^{n}}{9 ^{n}}

Apply Series Ratio Test:

converges

= \frac{1}{3} converges

= converges

\frac{d}{d x} (32 cos (x))

n = 1 \sum \infty \frac{x ^{n}}{1 6 ^{n}}

f (x) = - \frac{4 x}{x ^{2} + 1}, at x = 1

\frac{d y}{d x}, x^{\frac{2}{3}} + y^{\frac{2}{3}} = 20

y^{^{'''}} + 7 y^{^{''}} + 10 y^{^{'}} = 0

What is the sum from n=1 to infinity of ((-5)^n)/(3^{2n+1)} ?
The sum from n=1 to infinity of ((-5)^n)/(3^{2n+1)} is converges