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Is this power series a convergent series?

chamilka

New member
Jul 3, 2012
9
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!

gif2.gi.gif

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

Thank you for your kind support!!
 
Last edited by a moderator:

daigo

Member
Jun 27, 2012
60
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

Well-known member
Jan 26, 2012
890
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

New member
Jul 3, 2012
9
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..
 

daigo

Member
Jun 27, 2012
60
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

Indicium Physicus
Staff member
Jan 26, 2012
4,192
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

Well-known member
Jan 26, 2012
890
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

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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 here, 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

New member
Jul 3, 2012
9
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)