Mark M. answered • 12d

Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.

∑_{n=2 to ∞} [1 / √(n^{3} - 2n^{2} + 5n - 7]

Try the limit comparison test with a_{n} = 1 / √(n^{3} - n^{2} + 5n - 7) and b_{n} = 1 / √n^{3} = 1/n^{3/2}

∑ _{(n=2 to∞)} b_{n }converges since it is the tail end of the p-series with p = 3/2 > 1

a_{n} / b_{n} = √ [n^{3} / (n^{3} - n^{2} + 5n - 7)] = √ [ 1 / (1 - 1/n + 5/n^{2} - 7/n^{3})]

lim_{n→∞} [a_{n} / b_{n} ] = 1, so ∑_{(n=2 to ∞)} a_{n} also converges.