What is the sum from k=0 to infinity of k/(3k+1) ?

The sum from k=0 to infinity of k/(3k+1) is diverges

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sum from k=0 to infinity of k/(3k+1)

k = 0 \sum \infty \frac{k}{3 k + 1}

diverges

Solution steps

k = 0 \sum \infty \frac{k}{3 k + 1}

Apply Series Divergence Test:

diverges

= diverges

n = 0 \sum \infty (\frac{8}{7})^{n}

x = 0 \sum \infty \frac{x}{2 x + 1}

n = 0 \sum \infty arctan (\frac{π}{4})

k = 1 \sum \infty k e^{- 4 k}

n = 0 \sum \infty 10 ln (8 - n)

What is the sum from k=0 to infinity of k/(3k+1) ?
The sum from k=0 to infinity of k/(3k+1) is diverges