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

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

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

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

diverges

Solution steps

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

Simplify

= n = 1 \sum \infty 1

Every infinite sum of a non-zero constant diverges

= diverges

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

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

n = 0 \sum \infty n^{- 2}

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

n = 0 \sum \infty \frac{3}{n ^{2} + n - 2}

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