What is the sum from n=0 to infinity of (5/4)^{n-1} ?

The sum from n=0 to infinity of (5/4)^{n-1} is diverges

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sum from n=0 to infinity of (5/4)^{n-1}

n = 0 \sum \infty (\frac{5}{4})^{n - 1}

diverges

Solution steps

n = 0 \sum \infty (\frac{5}{4})^{n - 1}

Apply Series Ratio Test:

diverges

= diverges

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

k = 5 \sum \infty \frac{1}{( k - 4 ) ^{5}}

k = 0 \sum \infty (- \frac{1}{8})^{k}

n = 1 \sum \infty 1 - (- 1)^{n}

n = 0 \sum \infty cos (\frac{18}{n})

What is the sum from n=0 to infinity of (5/4)^{n-1} ?
The sum from n=0 to infinity of (5/4)^{n-1} is diverges