What is the sum from n=1 to infinity of (12^n)/(n+1) ?

The sum from n=1 to infinity of (12^n)/(n+1) is diverges

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sum from n=1 to infinity of (12^n)/(n+1)

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

diverges

Solution steps

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

Apply Series Ratio Test:

diverges

= diverges

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

n = 1 \sum \infty \frac{1}{( 4 n + 5 ) ^{3}}

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

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

k = 2 \sum \infty k e^{- 3 k^{2}}

What is the sum from n=1 to infinity of (12^n)/(n+1) ?
The sum from n=1 to infinity of (12^n)/(n+1) is diverges