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

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

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

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

converges

Solution steps

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

Apply Series Comparison Test:

converges

= converges

n = 1 \sum \infty \frac{8}{( - 3 ) ^{n}}

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

n = 5 \sum \infty 2 (- \frac{1}{4})^{n}

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

n = 1 \sum \infty \frac{1}{8 n + 4}

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