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I was tasked with determining whether the following series converge:

$$\displaystyle\sum _{n=2}^{\infty }\:\frac{\cos\left(n\right)}{n^3-n}$$

$$\displaystyle\sum _{n=2}^{\infty }\:\frac{1}{n\left(\ln\left(n\right)\right)^3}$$

In the first, I tried employing the integral test which failed, specifically because it is around 2 not 0 as we see in the plain theory. I couldn't find the solution on the given sheet to see whether I was doing it correctly or not. I was suggested that the Maclaurin series might be of use here, but I'm not sure how to employ it.

With the second, I know that it indeed converges, but the solution is so complicated with the Integral Tests that I'm having serious trouble with it.

Arturo Magidin
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Elliot
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4 Answers4

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For the first one, you can use the comparison test, since$$n>1\implies\frac{\left|\frac{\cos(n)}{n^3-n}\right|}{\frac1{n^3}}\leqslant\frac{n^3}{n^3-n}\to1.$$

In the case of the second one, the integral test works just fine: since$$\int\frac1{x\log^3(x)}\,\mathrm dx=-\frac1{2\log^2(x)},$$you have$$\int_2^\infty\frac1{x\log^3(x)}\,\mathrm dx=\left[-\frac1{2\log^2(x)}\right]_{x=2}^{x=\infty}=\frac1{2\log^2(2)}.$$

José Carlos Santos
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For the first one, use the comparison test as Mark Viola suggested: $$\left|\frac{\cos n}{n^3-n}\right|\le \frac{1}{n^3-n} \le \frac{1}{\frac{1}{2}n^3} = \frac{2}{n^3}$$ Therefore, the series converges.

For the second one, it is a bit easier with the integral test: $$\sum _{n=2}^{\infty }\:\frac{1}{n\ln^3\left(n\right)} \iff \int_2^\infty \frac{dx}{x\ln^3x} = -\frac{1}{2\ln^2x}\bigg|_2^\infty = 0 + \frac{1}{2\ln^22}$$

VIVID
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    @MarkViola Yes but no problem, let's provide the OP with variety of approaches then :) – VIVID Apr 14 '21 at 17:13
  • First, the integral test doesn't even apply since the integrand is NOT monotonic due to the presence of the term $\cos(n)$! But, my point was that Dirichlet's test is likely beyond the level of the OP's courses. So, why even mention it? Rather, use an approach that the OP has already learned (e.g., p-test, comparison test, etc.). – Mark Viola Apr 14 '21 at 17:33
  • @MarkViola Integrand does not need to be monotonic but only one of the functions in the product. In this case, we choose $f(x) = \cos x$ which has a bounded antiderivative and $g(x) = \frac{1}{x^3-x}$ which is eventually decreasing and tends to $0$. – VIVID Apr 14 '21 at 17:47
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    In order to apply the INTEGRAL TEST (not Dirichlet's) in general, the summand needs to be monotonic. – Mark Viola Apr 14 '21 at 17:52
  • @MarkViola Thank you for pointing it out. – VIVID Apr 14 '21 at 17:58
  • You're welcome. My pleasure. – Mark Viola Apr 14 '21 at 17:58
  • So, please modify accordingly as the current answer is wrong and misleading. – Mark Viola Apr 14 '21 at 17:58
  • @MarkViola I was on my way. Done :) – VIVID Apr 14 '21 at 18:00
  • Grazie mille, everyone! I think I managed to reach a conclusion now. – Elliot Apr 14 '21 at 19:03
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For the first one: $$\frac{\cos(n)}{n^3-n}\le \frac{1}{n^3-n}\approx\frac{1}{n^3}$$ which converges as long as the index starts at $1$ or higher, which it does.

For the second one: $$\frac{1}{(n\ln n)^3}$$ we know that: $$\ln(n)>1\,\,\,,n>3$$ and so: $$\sum_{n=3}^\infty\frac{1}{(n\ln(n))^3}\le\sum_{n=3}^\infty\frac1{n^3}<\infty$$ now just check that the $n=2$ term converges: $$0<\frac{1}{(2\ln(2))^3}=\frac{1}{\ln^3(4)}<1$$ since $$\ln(4)>\ln(e)=1\therefore \ln^3(4)>1$$

Henry Lee
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For the first one,

$\frac{cos(n)}{n^3 -n}$ $\lt$ $\frac{1}{n^3 -n}$

$\frac{1}{n^3 -n}$ = $\frac{1}{n(n^2-1)}$ = $\frac{1}{n(n+1)(n-1)}$

$\Rightarrow$ $\frac{1}{n^3 -n}$ = $\frac{1}{(n-1)n(n+1)}$ = $\frac{1}{n-1} - \frac{2}{n} + \frac{1}{n+1} $

$\Rightarrow$ $\sum_2^n \frac{cos(n)}{n^3 -n}$ $\lt$ $\sum_2^n \frac{1}{n^3 -n}$ = $\sum_2^n (\frac{1}{n-1} - \frac{2}{n} + \frac{1}{n+1}) $ = $\frac{1}{2}$

Second one follows from integral test!

Vijay Honkalaskar
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