What is the sum from n=0 to infinity of 5(2/2)^9 ?

The sum from n=0 to infinity of 5(2/2)^9 is diverges

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sum from n=0 to infinity of 5(2/2)^9

n = 0 \sum \infty 5 (\frac{2}{2})^{9}

diverges

Solution steps

n = 0 \sum \infty 5 (\frac{2}{2})^{9}

Every infinite sum of a non-zero constant diverges

= diverges

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

x = 1 \sum \infty \frac{2}{x ^{2} + 4 x + 3}

n = 0 \sum \infty \frac{1}{cos ( \frac{1}{n} )}

n = 1 \sum \infty \frac{1}{4 n ^{2} - 81}

n = 0 \sum \infty \frac{1}{n ^{\frac{1}{6}}}

What is the sum from n=0 to infinity of 5(2/2)^9 ?
The sum from n=0 to infinity of 5(2/2)^9 is diverges