What is the sum from n=1 to infinity of 7^n ?

The sum from n=1 to infinity of 7^n is diverges

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sum from n=1 to infinity of 7^n

n = 1 \sum \infty 7^{n}

diverges

Solution steps

n = 1 \sum \infty 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 7^{n} - 7^{0}

n = 0 \sum \infty 7^{n} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

n = 1 \sum \infty e^{- 7 n}

n = 1 \sum \infty 6 e^{- n}

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

k = 5 \sum \infty 3 (\frac{1}{3})^{3 k}

n = 2 \sum \infty (\frac{1}{5})^{2 n}

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