Convergence of ∑ n = 1 ∞ n + 5 n 7 + n 2 − n 3

Question

Convergence of      ∑          n      =      1              ∞                  n      +      5                                n          7                +                  n          2                −        n            3

Ethen Valentine · Accepted Answer

You almost did the right thing at the start.
Compare with the series whose terms are the "important' terms in the expression

\frac{n + 5}{\sqrt[3]{n^{7} + n^{2} - n}} .

This leads you to compare with the series

\sum_{n = 1}^{\infty} \frac{1}{n^{4 / 3}} .

(do not compare with the series whose terms are

1 / n^{2}

).
Note that

\sum_{n = 1}^{\infty} \frac{1}{n^{4 / 3}}

is a convergent p series (the p-series

\sum_{n = 1}^{\infty} \frac{1}{n^{p}}

converges if and only if

p > 1

). So the original series will converge, if the Limit Comparison Test applies. But, of course, we need to check if the Limit Comparison Test applies; so, we take the limit

\begin{aligned} lim_{n \to \infty} \frac{\frac{n + 5}{\sqrt[3]{n^{7} + n^{2} - n}}}{\frac{1}{n^{4 / 3}}} & = lim_{n \to \infty} \frac{n + 5}{\sqrt[3]{n^{7} + n^{2} - n}} \cdot n^{4 / 3} \\ = lim_{n \to \infty} \frac{n^{7 / 3} + 5 n^{4 / 3}}{\sqrt[3]{n^{7} (1 + \frac{1}{n^{5}} - \frac{1}{n^{6}})}} \\ = lim_{n \to \infty} \frac{n^{7 / 3} (1 + \frac{5}{n^{3 / 3}})}{n^{7 / 3} \sqrt[3]{1 + \frac{1}{n^{5}} - \frac{1}{n^{6}}}} \\ = lim_{n \to \infty} \frac{1 + \frac{5}{n^{3 / 3}}}{\sqrt[3]{1 + \frac{1}{n^{5}} - \frac{1}{n^{6}}}} \\ = 1. \end{aligned}

Since limit value is 1, the Limit Comparison Test indeed applies; and thus

\sum_{n = 1}^{\infty} \frac{n + 5}{\sqrt[3]{n^{7} + n^{2} - n}}

is a convergent series.

Convergence of <munderover> ∑ <mrow class="MJX-TeXAtom-ORD"> n

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