Does this series converge or diverge?

Notice that for large k, 1/√(2k^2 - k) behaves similarly to 1/√(2k^2) = 1/(k√2) (since 2k^2 dominates k as k gets large). So, by dropping the superfluous constant, we should compare the given series to 1/k by the Limit Comparsion Test.

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We know that Σ 1/k (from k=1 to infinity) diverges (as it is the divergent harmonic series). If we let a(k) = 1/k and b(k) = 1/√(2k^2 - k), then we have:

lim (k-->infinity) a(k)/b(k) = √2,

which is positive and finite limit. Thus, by the Limit Comparsion Test, Σ a(k) and Σ b(k) either both converge or both diverge; however, we know that Σ a(k) diverges, so Σ b(k) must also diverge.

Therefore, the series diverges by the Limit Comparsion Test.

If k is a positive integer, then (2k^2 - k) will always grow as k grows. You can factor it to (k)(2k - 1) to see that.

If (2k^2 - k) grows, then so does its square root.

If sqrt(2k^2 - 2k) grows, then its reciprocal 1/sqrt(2k^2 - k) gets SMALLER. You can test increasing values of k if you like; you'll see that this is true.

Since each term of the series gets smaller, the series converges.