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

The sum from n=1 to infinity}3^{-n of+5^{-n} is 3/4

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sum from n=1 to infinity}3^{-n of+5^{-n}

n = 1 \sum \infty 3^{- n} + 5^{- n}

\frac{3}{4}

Solution steps

n = 1 \sum \infty 3^{- n} + 5^{- n}

Apply the Sum Rule:

\sum a_{n} + b_{n} = \sum a_{n} + \sum b_{n}

= n = 1 \sum \infty 3^{- n} + n = 1 \sum \infty 5^{- n}

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

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

= \frac{1}{2} + \frac{1}{4}

Simplify

\frac{1}{2} + \frac{1}{4} : \frac{3}{4}

= \frac{3}{4}

n = 0 \sum \infty \frac{8}{5 - n}

n = 1 \sum \infty (cos (18))^{n}

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

n = 0 \sum \infty 8^{- n}

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

What is the sum from n=1 to infinity}3^{-n of+5^{-n} ?
The sum from n=1 to infinity}3^{-n of+5^{-n} is 3/4