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Gauss Test states that:Let $$\sum_{n=1}^\infty u_n$$ be a positive term series and let there exist two positive numbers $\rho , \alpha $ and a bounded sequence $\langle a_n\rangle $ such that

$ \frac{u_n}{u_{n+1}} = 1 + \frac{\rho}{n} + \frac{a_n}{n^{1+\alpha}}$ . Then series $$\sum_{n=1}^\infty u_n$$ converges if $\rho > 1$ and diverges if $\rho \le 1$

There is a corollary given for this which I am required to proof but while proving it got some doubts. Please look into it.

Corollary: If there exists $\alpha >0$ such that

$$\lim_{n\to \infty} [n^{\alpha} ( n (\frac{u_n}{u_{n+1}}-1)) ] $$

exists finitely, then the series $$\sum_{n=1}^\infty u_n$$ is divergent.

My attempt: From gauss test

$ [n^{\alpha} ( n (\frac{u_n}{u_{n+1}}-1)) ] $ = $ n^{\alpha} \rho + a_n $

Also, $|a_n| \le M$, where $M$ is a real constant

Therefore,

$ [n^{\alpha} ( n (\frac{u_n}{u_{n+1}}-1)) ] $ = $ n^{\alpha} \rho + a_n \le n^{\alpha} \rho + M $

Taking limit n tends to $\infty$ on both sides and analysing RHS,

$\lim_{n\to \infty} n^{\alpha} \rho + M $

If this limit exists finitely, then $\rho$ has to be less than or equal to 1 i.e. I have to show that $\rho$ cannot be greater than 1.

Doubt 1: For any $\alpha \gt 0$,

$\lim_{n\to \infty} n^{\alpha} = \infty $ always ?

Doubt 2: How the finite existence of limit put the condition on $\rho$ that it cannot be greater than 1? I need the help with the proof.

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  • $\begingroup$ I don't think you stated Gauss test correctly. $\endgroup$ – user10354138 Aug 1 at 14:57
  • $\begingroup$ It is stated correctly. $\endgroup$ – Shashank Dwivedi Aug 1 at 15:15
  • $\begingroup$ No it is not. If it is, then $\frac{u_n}{u_n+1}<1$ since $u_n>0$, so there are no $\rho>0$ with property $\frac{u_n}{u_n+1}=1+\rho n^{-1}+O(n^{-1-\alpha})$ making the test useless. $\endgroup$ – user10354138 Aug 1 at 15:19
  • $\begingroup$ I guess the problem is $u_{n+1}$ has been written as $u_n +1$ $\endgroup$ – FormulaWriter Aug 1 at 15:23
  • $\begingroup$ Corrected. Thank you for observing that. $\endgroup$ – Shashank Dwivedi Aug 1 at 15:25
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If $ 0<\alpha \le 1 $, then $\lim_{n\to \infty} [n^{\alpha} ( n (\frac{u_n}{u_{n+1}}-1)) ] $ becomes $0$ only if $\rho=0 $.

Otherwise, for $\rho > 0 $ , $\lim_{n\to \infty} [n^{\alpha} ( n (\frac{u_n}{u_{n+1}}-1)) ] $ goes to $\infty $, and for $\rho < 0 $,$\lim_{n\to \infty} [n^{\alpha} ( n (\frac{u_n}{u_{n+1}}-1)) ] $ goes to $-\infty $

And if, $ \alpha \gt 1 $, then anything can be happened. $\lim_{n\to \infty} [n^{\alpha} ( n (\frac{u_n}{u_{n+1}}-1)) ] $ can be either $0$ or $\infty$ or $-\infty $ or undefined.

For example, take the series $\sum (1-\frac{1}{n}) $ , and take $ \alpha = \frac{5}{2} $, then $\lim_{n\to \infty} [n^{\alpha} ( n (\frac{u_n}{u_{n+1}}-1)) ] $ becomes $\infty - \infty $ , which is clearly undefined.

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