What is the sum from n=1 to infinity of (8^n)/(n+1) ?

The sum from n=1 to infinity of (8^n)/(n+1) is diverges

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sum from n=1 to infinity of (8^n)/(n+1)

n = 1 \sum \infty \frac{8 ^{n}}{n + 1}

diverges

Solution steps

n = 1 \sum \infty \frac{8 ^{n}}{n + 1}

Apply Series Ratio Test:

diverges

= diverges

n = 0 \sum \infty \frac{1}{( n + 2 ) n !}

n = 1 \sum \infty \frac{2 n + 1}{6 ^{n}}

n = 0 \sum \infty \frac{4}{3 - n}

n = 0 \sum \infty \frac{1}{n ^{4} + n ^{2}}

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

What is the sum from n=1 to infinity of (8^n)/(n+1) ?
The sum from n=1 to infinity of (8^n)/(n+1) is diverges