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#### OhMyMarkov

##### Member

- Mar 5, 2012

- 83

Hello Everyone!

Following my earlier post, I've been trying to prove something that uses the convergence of a power series (which I will reveal should I succeed to complete the proof).

Suppose the series $\sum _{n=0} ^{\infty} c_n x^n$ converges for all

I am reluctant to write the following:

|$\sum _{n=0} ^{\infty} c_n x^{n-1}|$ $<|\sum _{n=0} ^{\infty} c_n x^n|$

because $x$ can be bigger than 1.

However, with a simple change of variables $k=n-1$, I obtain the following:

$S=\sum _{k=0} ^{\infty} c_k k x^k$ is a series that converges for:

$|x|<R'=\displaystyle \frac{1}{\lim \sup \sqrt[k]{c_k k}}=\frac{1}{\lim \sup \sqrt[k]{k}}$ $\cdot \displaystyle \frac{1}{\lim \sup \sqrt[k]{c_k}}=1.R = R$

Hence, $S$ converges...

Is the proof correct?

Following my earlier post, I've been trying to prove something that uses the convergence of a power series (which I will reveal should I succeed to complete the proof).

Suppose the series $\sum _{n=0} ^{\infty} c_n x^n$ converges for all

**real**$|x|<R$ where $R$ is the radius of convergence. Now, does $\sum _{n=1} ^{\infty} c_n nx^{n-1}$ converge as well?I am reluctant to write the following:

|$\sum _{n=0} ^{\infty} c_n x^{n-1}|$ $<|\sum _{n=0} ^{\infty} c_n x^n|$

because $x$ can be bigger than 1.

However, with a simple change of variables $k=n-1$, I obtain the following:

$S=\sum _{k=0} ^{\infty} c_k k x^k$ is a series that converges for:

$|x|<R'=\displaystyle \frac{1}{\lim \sup \sqrt[k]{c_k k}}=\frac{1}{\lim \sup \sqrt[k]{k}}$ $\cdot \displaystyle \frac{1}{\lim \sup \sqrt[k]{c_k}}=1.R = R$

Hence, $S$ converges...

Is the proof correct?

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