What is the sum from n=1 to infinity of (3^n)/(n+1) ?

The sum from n=1 to infinity of (3^n)/(n+1) is diverges

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sum from n=1 to infinity of (3^n)/(n+1)

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

diverges

Solution steps

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

Apply Series Ratio Test:

diverges

= diverges

k = 1 \sum \infty ln (\frac{k}{3 k + 1})

n = 0 \sum \infty \frac{n}{8 n + 1}

n = 0 \sum \infty sin (\frac{n}{2})

n = 0 \sum \infty \frac{1}{1 - 3 n}

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

What is the sum from n=1 to infinity of (3^n)/(n+1) ?
The sum from n=1 to infinity of (3^n)/(n+1) is diverges