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.

Question

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.

Answer 1

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!