# Prove that $\sum_{n=0}^\infty c_nx^n$ and $\sum_{n=0}^\infty n\log n c_nx^{n+3}$ have same radius of convergence.

Prove that $$\sum_{n=0}^\infty c_nx^n$$ and $$\sum_{n=0}^\infty n\log n c_nx^{n+3}$$ have same radius of convergence.

Now for $$\sum_{n=0}^\infty c_nx^n$$ the radius of convergence is $$\frac{1}{\limsup\left|\frac {c_{n+1}}{c_{n}}\right|}$$. Again, for the series $$\sum_{n=0}^\infty n\log n c_nx^{n+3}$$ we have $$\limsup\frac{(n+1)\log (n+1) c_{n+1}|x|^{n+4}}{n\log n c_n|x|^{n+3}}<1$$ this will imply that $$|x|<\frac{1}{\limsup\left|\frac {c_{n+1}}{c_{n}}\right|\limsup\left|\frac {{n+1}}{{n}}\right|\limsup\left|\frac {\log {n+1}}{\log {n}}\right|}.$$

Now $$\limsup\left|\frac {{n+1}}{{n}}\right|=1$$, what should I do with $$\limsup\left|\frac {\log {n+1}}{\log {n}}\right|$$?

Please help me from here. Also, let me know if there is any mistake in this attempt and also if you know any other methods.

• Note that $\lim_n \dfrac{\ln (n+1)}{\ln n}=1.$
– mfl
Commented Dec 26, 2020 at 20:29
• Your formula for the radius of convergence is wrong. It's $$\frac{1}{R} = \limsup_{n \to \infty} \lvert c_n\rvert^{1/n}\,.$$ In case the limit exists, we also have $$\frac{1}{R} = \lim_{n \to \infty} \frac{\lvert c_{n+1}\rvert}{\lvert c_n\rvert}\,,$$ but if the limit doesn't exist, one typically cannot replace $\lim$ with $\limsup$ there. Commented Dec 26, 2020 at 20:36
• @DanielFischer I understand. Is there any other way to solve the problem then? Commented Dec 26, 2020 at 20:47
• I think using $\limsup c_n$, we can prove it as well. Commented Dec 26, 2020 at 20:48
• But I am having problem with $\limsup[(n-3)\log(n-3) |c_{n-3}|]^{\frac 1 n}=R_2$ the radius of convergence of the 2nd series. Commented Dec 26, 2020 at 20:57

Supposing the limit exists. Be $$R$$ the radius of convergence, this is:

$$1/R = \limsup\limits_{n\to\infty} \sqrt[n]{|c_n|}$$

By according to Cauchy, Hadamard's Theorem. But it's evident that:

$$\limsup\limits_{n\to\infty} \sqrt[n]{|c_n|} \leq \limsup\limits_{n\to\infty} \sqrt[n]{\frac{|c_{n+1}|}{c_n}}$$

Then, we have that

$$\limsup\limits_{n\to\infty} \sqrt[n]{|c_n|} \leq \limsup\limits_{n\to\infty} \sqrt[n]{\frac{|c_{n+1}|}{c_n}}$$

It's clear that,

$$\limsup\limits_{n\to\infty} \sqrt[n]{\frac{|c_{n+1}|}{c_n}} = \lim\limits_{n\to\infty} \frac{(n+1) \log (n+1)}{n \log n} = 1$$,

we got this result applying L'Hopital's Rule.

this happens because whenever a limit does exist, the lim sup and lim inf also.

There is also something we should notice is that:

Suppose we have a series $$\sum_{i = 0}^{\infty} c_n x^{ap + n}$$ converges to the same radius of convergence r of $$x^{ap} \sum_{i = 0}^{\infty} c_n x^{n}$$. And this is important because of our $$x^3$$ in the problem. This way it doesn't make difference at all!