Is this power series a convergent series?

[chamilka]

Hi everyone!
I have got this series in a part of my research. I need to check if this is a convergent series and if so, what is the radius of the convergence?

Here is the series..
\[\sum_{i=0}^{\infty}(-1)^{i}{b-1\choose i}B(y+ac+ci,\,n-y+1)\]

Sorry if my LateX code is not visible( I am currently learning LaTeX as said in the Forum rules), see the attached image below!

View attachment 225

Here, a,b and c are any three positive real numbers and y=0,1,2,...n

Thank you for your kind support!

 

[hatelove]

Hi chamilka, just wrap the LaTeX code in [TEX] [/ TEX] tags like so:

[TEX]sum_{i=0}^{\infty}(-1)^{i}{b-1\choose i}B(y+ac+ci,\,n-y+1)[ /TEX]

(Just remove the space before the forward slash in the closing tex tag)

 

[CaptainBlack]

Hi chamilka, just wrap the LaTeX code in [TEX] [/ TEX] tags like so:

[TEX]sum_{i=0}^{\infty}(-1)^{i}{b-1\choose i}B(y+ac+ci,\,n-y+1)[ /TEX]

(Just remove the space before the forward slash in the closing tex tag)

Wrong!, on this site wrap the LaTeX with either \$ .. \$ or \$\$ .. \$\$ tags.

CB

 

[chamilka]

Hi chamilka, just wrap the LaTeX code in [TEX] [/ TEX] tags like so:

[TEX]sum_{i=0}^{\infty}(-1)^{i}{b-1\choose i}B(y+ac+ci,\,n-y+1)[ /TEX]

(Just remove the space before the forward slash in the closing tex tag)

and

Wrong!, on this site wrap the LaTeX with either \$ .. \$ or \$\$ .. \$\$ tags.

CB

Thank you daigo and CaptainBlack for your kind LateX teaching.. Special thanks to CaptainBlack who just edited my post..

 

[hatelove]

Wrong!, on this site wrap the LaTeX with either \$ .. \$ or \$\$ .. \$\$ tags.

CB

I've never done this before, I always used the [TEX] tags because I'm so used to it...but then how do you write dollar signs?

test i made $25 today and $30 yesterday test

 

[Ackbach]

I've never done this before, I always used the [TEX] tags because I'm so used to it...but then how do you write dollar signs?

You escape them thus: $\$5,600$. Use a backslash before the dollar sign inside a math environment.

 

[CaptainBlack]

I've never done this before, I always used the [TEX] tags because I'm so used to it...but then how do you write dollar signs?

You look at how I got the dollar signs to display in the text you quoted.

CB

 

[Sudharaka]

Hi everyone!
I have got this series in a part of my research. I need to check if this is a convergent series and if so, what is the radius of the convergence?

Here is the series..
\[\sum_{i=0}^{\infty}(-1)^{i}{b-1\choose i}B(y+ac+ci,\,n-y+1)\]

Sorry if my LateX code is not visible( I am currently learning LaTeX as said in the Forum rules), see the attached image below!

View attachment 225

Here, a,b and c are any three positive real numbers and y=0,1,2,...n

Thank you for your kind support!

Hi chamilka, :)

Firstly I think you should review what a power series is. The given series is not a power series and the radius of convergence is defined only for power series.

\[\sum_{i=0}^{\infty}(-1)^{i}{b-1\choose i}B(y+ac+ci,\,n-y+1)\]

Using the method that we have used http://www.mathhelpboards.com/threads/1358-Chamilka-s-Question-from-Math-Help-Forum?p=6494#post6494, this series can be expressed as the following integral.

\begin{eqnarray}

\sum_{i=0}^{\infty}(-1)^{i}{b-1\choose i}B(y+ac+ci,\,n-y+1)&=&\int_{0}^{1}x^{y+ac-1}(1-x)^{n-y}(1-x^c)^{b-1}\,dx\\

&=&\int_{0}^{1}x^{y+ac-1}\left(\sum_{j=0}^{n-y}(-1)^{j}{n-y\choose j}x^{j}\right)(1-x^c)^{b-1}\,dx\\

&=&\sum_{j=0}^{n-y}\int_{0}^{1}x^{y+ac+j-1}(-1)^{j}{n-y\choose j}(1-x^c)^{b-1}\,dx\\

&=&\sum_{j=0}^{n-y}\int_{0}^{1}(-1)^{j}{n-y\choose j}(x^c)^{\left(\frac{y}{c}+a+\frac{j}{c}-\frac{1}{c}\right)}(1-x^c)^{b-1}\,dx\\

&=&\sum_{j=0}^{n-y}(-1)^{j}{n-y\choose j}\int_{0}^{1}(x^c)^{\left(\frac{y}{c}+a+\frac{j}{c}-\frac{1}{c}\right)}(1-x^c)^{b-1}\,dx\\

\end{eqnarray}

Let, \(\displaystyle x=z^{\frac{1}{c}}\Rightarrow dx=\frac{1}{c}z^{\frac{1}{c}-1}\,dz\)

\begin{eqnarray}

\therefore\sum_{i=0}^{\infty}(-1)^{i}{b-1\choose i}B(y+ac+ci,\,n-y+1)&=&\sum_{j=0}^{n-y}(-1)^{j}{n-y\choose j}\int_{0}^{1}z^{\left(\frac{y}{c}+a+\frac{j}{c}-\frac{1}{c}\right)}(1-z)^{b-1}\frac{1}{c}z^{\frac{1}{c}-1}\,dz\\

&=&\frac{1}{c}\sum_{j=0}^{n-y}(-1)^{j}{n-y\choose j}\int_{0}^{1}z^{\left(\frac{y}{c}+a+\frac{j}{c}-1\right)}(1-z)^{b-1}\,dz\\

&=&\frac{1}{c}\sum_{j=0}^{n-y}(-1)^{j}{n-y\choose j}B\left(\frac{y}{c}+a+\frac{j}{c},\,b\right)

\end{eqnarray}

\(\sum_{j=0}^{n-y}(-1)^{j}{n-y\choose j}B\left(\frac{y}{c}+a+\frac{j}{c},\,b\right)\) is a finite series since, \(n-y\in\mathbb{Z}^{+}\cup\{0\}\)

Therefore, \(\sum_{i=0}^{\infty}(-1)^{i}{b-1\choose i}B(y+ac+ci,\,n-y+1)\) can be written as a finite series and hence it is convergent.

Kind Regards,
Sudharaka.

 

[chamilka]

Hi chamilka, :)

Firstly I think you should review what a power series is. The given series is not a power series and the radius of convergence is defined only for power series.
.......
\(\sum_{j=0}^{n-y}(-1)^{j}{n-y\choose j}B\left(\frac{y}{c}+a+\frac{j}{c},\,b\right)\) is a finite series since, \(n-y\in\mathbb{Z}^{+}\cup\{0\}\)

Therefore, \(\sum_{i=0}^{\infty}(-1)^{i}{b-1\choose i}B(y+ac+ci,\,n-y+1)\) can be written as a finite series and hence it is convergent.

Kind Regards,
Sudharaka.

Thank you very much Sudharaka. I got some idea about the power series and radius of convergence from the wiki articles and from the way you explained that my infinite series is in deed a finite series I got answer to my question without troubling more about PS and RoC.
I honestly express my gratitude.. (Handshake)