What is the sum from n=1 to infinity of (12/7)^n ?

The sum from n=1 to infinity of (12/7)^n is diverges

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sum from n=1 to infinity of (12/7)^n

n = 1 \sum \infty (\frac{12}{7})^{n}

diverges

Solution steps

n = 1 \sum \infty (\frac{12}{7})^{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{12}{7})^{n} - (\frac{12}{7})^{0}

n = 0 \sum \infty (\frac{12}{7})^{n} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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

n = 4 \sum \infty \frac{1}{n ^{5}}

n = 1 \sum \infty (- \frac{5}{8})^{n - 1}

n = 0 \sum \infty e^{- 17 n}

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

What is the sum from n=1 to infinity of (12/7)^n ?
The sum from n=1 to infinity of (12/7)^n is diverges