What is the sum from n=1 to infinity of (11/5)^n ?

The sum from n=1 to infinity of (11/5)^n is diverges

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sum from n=1 to infinity of (11/5)^n

n = 1 \sum \infty (\frac{11}{5})^{n}

diverges

Solution steps

n = 1 \sum \infty (\frac{11}{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{11}{5})^{n} - (\frac{11}{5})^{0}

n = 0 \sum \infty (\frac{11}{5})^{n} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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

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

n = 1 \sum \infty (\frac{1}{10})^{n - 1}

n = 1 \sum \infty 5 (\frac{3}{7})^{n}

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

What is the sum from n=1 to infinity of (11/5)^n ?
The sum from n=1 to infinity of (11/5)^n is diverges