What is the sum from n=1 to infinity of ((7pi)/8)^n ?

The sum from n=1 to infinity of ((7pi)/8)^n is diverges

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sum from n=1 to infinity of ((7pi)/8)^n

n = 1 \sum \infty (\frac{7 π}{8})^{n}

diverges

Solution steps

n = 1 \sum \infty (\frac{7 π}{8})^{n}

Apply Series Ratio Test:

diverges

= diverges

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

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

n = 0 \sum \infty \frac{4}{n ^{2} + 1}

n = 0 \sum \infty \frac{n ^{2}}{4 ^{n}}

n = 1 \sum \infty \frac{5}{6 ^{n - 1}}

What is the sum from n=1 to infinity of ((7pi)/8)^n ?
The sum from n=1 to infinity of ((7pi)/8)^n is diverges