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

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

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

n = 1 \sum \infty (\frac{π}{3})^{n}

diverges

Solution steps

n = 1 \sum \infty (\frac{π}{3})^{n}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

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

n = 0 \sum \infty (\frac{π}{3})^{n} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

k = 1 \sum \infty \frac{1}{k + 1}

n = 0 \sum \infty \frac{n}{9 + n ^{2}}

n = 0 \sum \infty 7 (\frac{2}{3})^{n + 2}

n = 1 \sum \infty \frac{e ^{\frac{8}{n}}}{n}

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

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