What is the sum from n=1 to infinity of ((3pi)/4)^n ?

The sum from n=1 to infinity of ((3pi)/4)^n is diverges

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sum from n=1 to infinity of ((3pi)/4)^n

n = 1 \sum \infty (\frac{3 π}{4})^{n}

diverges

Solution steps

n = 1 \sum \infty (\frac{3 π}{4})^{n}

Apply Series Ratio Test:

diverges

= diverges

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

n = 1 \sum \infty \frac{10}{n ^{2} + 9}

n = 0 \sum \infty (- \frac{3}{8})^{n}

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

n = 3 \sum \infty \frac{sin ( nπ )}{n}

What is the sum from n=1 to infinity of ((3pi)/4)^n ?
The sum from n=1 to infinity of ((3pi)/4)^n is diverges