What is the sum from k=3 to infinity of 1/(2k^2+k) ?

The sum from k=3 to infinity of 1/(2k^2+k) is converges

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sum from k=3 to infinity of 1/(2k^2+k)

k = 3 \sum \infty \frac{1}{2 k ^{2} + k}

converges

Solution steps

k = 3 \sum \infty \frac{1}{2 k ^{2} + k}

Apply Series Comparison Test:

converges

= converges

n = 1 \sum \infty \frac{9}{n ^{\frac{3}{7}}}

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

n = 1 \sum \infty (- e)^{- n}

n = 0 \sum \infty (- \frac{5}{7})^{n}

n = 1 \sum \infty \frac{9}{n}

What is the sum from k=3 to infinity of 1/(2k^2+k) ?
The sum from k=3 to infinity of 1/(2k^2+k) is converges