What is the sum from n=1 to infinity of n/(13n+4) ?

The sum from n=1 to infinity of n/(13n+4) is diverges

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sum from n=1 to infinity of n/(13n+4)

n = 1 \sum \infty \frac{n}{13 n + 4}

diverges

Solution steps

n = 1 \sum \infty \frac{n}{13 n + 4}

Apply Series Divergence Test:

diverges

= diverges

n = 0 \sum \infty \frac{4}{5 - n}

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

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

n = 0 \sum \infty (1 + \frac{2}{5 n})^{2}

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

What is the sum from n=1 to infinity of n/(13n+4) ?
The sum from n=1 to infinity of n/(13n+4) is diverges