# Gauss's test and its corollary

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.

• I don't think you stated Gauss test correctly. – user10354138 Aug 1 at 14:57
• It is stated correctly. – Shashank Dwivedi Aug 1 at 15:15
• 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. – user10354138 Aug 1 at 15:19
• I guess the problem is $u_{n+1}$ has been written as $u_n +1$ – FormulaWriter Aug 1 at 15:23
• Corrected. Thank you for observing that. – Shashank Dwivedi Aug 1 at 15:25

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.