What is the sum from n=0 to infinity of (1/(1.02))^n ?

The sum from n=0 to infinity of (1/(1.02))^n is converges

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sum from n=0 to infinity of (1/(1.02))^n

n = 0 \sum \infty (\frac{1}{1.02})^{n}

converges

Solution steps

n = 0 \sum \infty (\frac{1}{1.02})^{n}

Apply Series Ratio Test:

converges

= converges

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

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

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

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

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

What is the sum from n=0 to infinity of (1/(1.02))^n ?
The sum from n=0 to infinity of (1/(1.02))^n is converges