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I've a following task

Check convergence of:

$$\sum_{n=2}^\infty (-1)^{\lceil 1+\sin^{2} n^{5} \rceil}\left(\frac{n^{2}+3n+10}{n^{2}+5n+17}\right)^{n^{2}(\sqrt{n+1}-\sqrt{n-1})}$$

My solution is:

$\lceil 1+\sin^{2} n^{5} \rceil$ is always 2, because $\forall {k\in \mathbb{Z}}$, $\sin^{2} n^{5}\neq k\Pi$, where $\Pi$ is irrational.

$$\sum_{n=2}^\infty \left(\frac{n^{2}+3n+10}{n^{2}+5n+17}\right)^{n^{2}(\sqrt{n+1}-\sqrt{n-1})}$$

Checking Cauchy test:

$$\lim_{n\rightarrow\infty} \sqrt[n]{\left(\frac{n^{2}+3n+10}{n^{2}+5n+17}\right)^{n^{2}(\sqrt{n+1}-\sqrt{n-1})}}=\lim_{n\rightarrow\infty}\left(\frac{n^{2}+3n+10}{n^{2}+5n+17}\right)^{n(\sqrt{n+1}-\sqrt{n-1})}$$

$$=\lim_{n\rightarrow\infty} \left(1+\frac{-2n-7}{n^{2}+5n+17}\right)^{n(\sqrt{n+1}-\sqrt{n-1})}=\lim_{n\rightarrow\infty}e^{\frac{-2n-7}{n^{2}+5n+17} n(\sqrt{n+1}-\sqrt{n-1})}=$$

$$=\lim_{n\rightarrow\infty}e^{\frac{(-2n-7)2n}{(n^{2}+5n+17)(\sqrt{n+1}+\sqrt{n-1})}}=-1^{-}$$

Because the series limit tends to 1 from bottom, the series converge. Is that a true assumption?

Michael Hardy
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Joggi
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1 Answers1

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According to your calculation, there is no indication of whether the series converges or diverges. ($|a_n^{1/n}|$ converges or diverges as $n \rightarrow \infty$ according to the limit being > or < 1, and is inconclusive if = 1.)

You made a mistake above:

$$\frac{n^2+3 n+10}{n^2+5 n+17} \approx 1 - \frac{2}{n} $$

so that

$$\left ( \frac{n^2+3 n+10}{n^2+5 n+17} \right )^{n^2 [\sqrt{n+1} - \sqrt{n-1}]} \approx \left ( 1 - \frac{2}{n} \right )^{n^{3/2}} \approx \exp{(-2\sqrt{n})} $$

as $n \rightarrow \infty$. The above convergence test would be inconclusive here, but since the behavior at infinity is a decaying exponential, we can say that the series converges.

Ron Gordon
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  • Thanks! Do You have any hints how to solve it correctly? – Joggi Jan 09 '13 at 14:20
  • Solve what correctly? I'm afraid I do not understand. I think the point was to show convergence, right? – Ron Gordon Jan 09 '13 at 14:27
  • That mean, how to formal prove this $$\left ( \frac{n^2+3 n+10}{n^2+5 n+17} \right )^{n^2 [\sqrt{n+1} - \sqrt{n-1}]} \approx \left ( 1 - \frac{2}{n} \right )^{n^{3/2}} \approx \exp{(-2\sqrt{n})}$$, so that I can better understand. If possible, please show it as inequality. – Joggi Jan 09 '13 at 14:29
  • Unfortunately, I do not have the time to work out as an inequality, but I can say that the approximation of the rational function begins by plugging a factor of $n^2$ out form numerator and denominator, as looking at the first correction beyond 1 as $n \rightarrow \infty$, i.e., a Taylor expansion in $1/n$. Then use $\lim_{n \rightarrow \infty} \left (1 - \frac{2}{n} \right )^n = \exp{(-2)}$. – Ron Gordon Jan 09 '13 at 14:33