What is the sum from n=1 to infinity of (7/5)^n ?

The sum from n=1 to infinity of (7/5)^n is diverges

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sum from n=1 to infinity of (7/5)^n

n = 1 \sum \infty (\frac{7}{5})^{n}

diverges

Solution steps

n = 1 \sum \infty (\frac{7}{5})^{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{7}{5})^{n} - (\frac{7}{5})^{0}

n = 0 \sum \infty (\frac{7}{5})^{n} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

n = 1 \sum \infty \frac{1}{12 n + 13}

n = 1 \sum \infty 5 n + 2

n = 0 \sum \infty \frac{3}{2} (\frac{1}{2})^{n}

n = 1 \sum \infty \frac{3}{( 4 n + 2 ) ^{3}}

k = 0 \sum \infty 2^{k}

What is the sum from n=1 to infinity of (7/5)^n ?
The sum from n=1 to infinity of (7/5)^n is diverges