What is the sum from n=0 to infinity of (3.5)^{-n} ?

The sum from n=0 to infinity of (3.5)^{-n} is 1.4

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sum from n=0 to infinity of (3.5)^{-n}

n = 0 \sum \infty (3.5)^{- n}

1.4

Solution steps

n = 0 \sum \infty 3. 5^{- n}

Simplify

3. 5^{- n} : \frac{1}{3. 5 ^{n}}

= n = 0 \sum \infty \frac{1}{3. 5 ^{n}}

Apply Series Geometric Test:

1.4

= 1.4

n = 1 \sum \infty \frac{10 n}{4 n + 1}

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

k = 1 \sum \infty 30 (0.1)^{k}

n = 1 \sum \infty \frac{8}{n} - \frac{8}{n ^{2}}

n = 1 \sum \infty (\frac{π}{n})^{n} \cdot n!

What is the sum from n=0 to infinity of (3.5)^{-n} ?
The sum from n=0 to infinity of (3.5)^{-n} is 1.4