If Σc_nx^n converges does Σc_n·nx^(n-1) converge as well?

[OhMyMarkov]

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 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?

 

[Opalg]

Re: If $\sum c_n x^n$ converges does $\sum c_n nx^{n-1}$ converge as well?

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 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?

Yes, the proof is correct*. The two power series always have the same radius of convergence.

* Edit. ... except that you should have used $|c_k|$ instead of $c_k$ (because otherwise you might not be able to take the $k$th root, if $c_k$ is negative).

 

[chisigma]

Re: If $\sum c_n x^n$ converges does $\sum c_n nx^{n-1}$ converge as well?

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 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?

In Your 'earlier post' it has benn demonstrated that if $\displaystyle f(x)= \sum_{n=0}^{\infty} a_{n}\ z^{n}$ converges absolutely for |x|<R, then f(x) is analytic for any |x|<R and its derivative is also analityc for |x|<R, so that f'(x) can be written as... $\displaystyle f^{\ '} (x)= \sum_{n=1}^{\infty} b_{n}\ x^{n-1}$ ... where... $\displaystyle b_{n}= n\ a_{n}$

Kind regards

$\chi$ $\sigma$

 

[OhMyMarkov]

Re: If $\sum c_n x^n$ converges does $\sum c_n nx^{n-1}$ converge as well?

To explain what I am trying to prove in these two posts:

If $\sum _{n=0} ^{\infty} c_n x^n$ converges for $\forall |x|<R$ to $f(x)$, show that $f(x)$ is continuous (this was these two posts).

@chisigma
My next step is to show that $f$ is differentiable with the derivative function exactly as the one you have written...

 

[Opalg]

Re: If $\sum c_n x^n$ converges does $\sum c_n nx^{n-1}$ converge as well?

To explain what I am trying to prove in these two posts:

If $\sum _{n=0} ^{\infty} c_n x^n$ converges for $\forall |x|<R$ to $f(x)$, show that $f(x)$ is continuous (this was these two posts).

@chisigma
My next step is to show that $f$ is differentiable with the derivative function exactly as the one you have written...

Since $\sum _{n=1} ^{\infty} nc_n x^{n-1}$ has the same radius of convergence $R$ as $\sum _{n=0} ^{\infty} c_n x^n$, it must converge (within that radius of convergence) to some function, say $g(x) = \sum _{n=1} ^{\infty} nc_n x^{n-1}.$ In fact, that series converges uniformly in any closed interval contained in $(-R,R)$. But a uniform limit of continuous functions is continuous and integrable, and can be integrated term by term. (Those are standard theorems that you should be able to find in any analysis text.) Therefore $\displaystyle\int g(x)\,dx = f(x)$ (plus a constant), from which it follows that $f'(x) = g(x).$

 

[chisigma]

Re: If $\sum c_n x^n$ converges does $\sum c_n nx^{n-1}$ converge as well?

A 'rigorous' statement of the problem requires the use of the so-called 'Cauchy Integral Formula' , according to which if f(z) is analytic inside and on a simple closed line C and z=a is inside C, then...

$\displaystyle f^{(n)} (a) = \frac{n!}{2\ \pi\ i}\ \oint_{C} \frac{f(z)}{(z-a)^{n+1}}\ dz$ (1)

If f(z) is analytic for z=a, then its Taylor expansion is...

$\displaystyle f(z)= a_{0} + a_{1}\ (z-a) + a_{2}\ (z-a)^{2} + ... = \sum_{k=0}^{\infty} a_{k}\ (z-a)^{k}$ (2)

Now if we integrate (2) and take into account that...

$\displaystyle \oint_{C} \frac{dz}{(z-a)^k} = \begin{cases}0 & \text{if } k \ne 1\\
2\ \pi\ i & \text{if } k=1 \end{cases}$ (3)

... setting n=0 in (1) we obtain...

$\displaystyle a_{k} = \frac{1}{2\ \pi\ i}\ \oint_{C} \frac{f(z)}{(z-a)^{k+1}}\ dz$ (3)

In similar way, writing...

$\displaystyle f^{\ '}(z)= b_{0} + b_{1}\ (z-a) + b_{2}\ (z-a)^{2} + ... = \sum_{k=0}^{\infty} b_{k}\ (z-a)^{k}$ (2)

... and setting n=1 in (1) we obtain...

$\displaystyle b_{k} = \frac{k+1}{2\ \pi\ i}\ \oint_{C} \frac{f(z)}{(z-a)^{k+2}}\ dz = (k+1)\ a_{k+1}$ (4)

Kind regards

$\chi$ $\sigma$

 

[Opalg]

Re: If $\sum c_n x^n$ converges does $\sum c_n nx^{n-1}$ converge as well?

A 'rigorous' statement of the problem requires the use of the so-called 'Cauchy Integral Formula' , according to which if f(z) is analytic inside and on a simple closed line C and z=a is inside C, then...

$\displaystyle f^{(n)} (a) = \frac{n!}{2\ \pi\ i}\ \oint_{C} \frac{f(z)}{(z-a)^{n+1}}\ dz$ (1)

It is certainly not necessary to use the heavy machinery of complex analysis to give a rigorous proof of this result about a function of a real variable. Standard results from real analysis (uniform convergence theorems, as in comment #5 above) are all that is needed.