What is the sum from n=1 to infinity of (9^n)/(n+1) ?

The sum from n=1 to infinity of (9^n)/(n+1) is diverges

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sum from n=1 to infinity of (9^n)/(n+1)

n = 1 \sum \infty \frac{9 ^{n}}{n + 1}

diverges

Solution steps

n = 1 \sum \infty \frac{9 ^{n}}{n + 1}

Apply Series Ratio Test:

diverges

= diverges

n = 0 \sum \infty (\frac{10}{11})^{n}

i = 0 \sum \infty \frac{1}{2 ^{i}}

n = 3 \sum \infty \frac{4}{7 ^{n}}

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

n = 70 \sum \infty \frac{3}{n ^{2} + 2 n}

What is the sum from n=1 to infinity of (9^n)/(n+1) ?
The sum from n=1 to infinity of (9^n)/(n+1) is diverges