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Infinite series sum

jacks

Well-known member
Apr 5, 2012
226
If $\displaystyle y=\frac{3}{4}+\frac{3*5}{4*8}+\frac{3*5*7}{4*8*12}+.....\infty$. Then $y^2+2y = $
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: infinite series sum

Can you tell us what your efforts have been and where you are stuck? Posting a question without any work does not give our helpers a good place to begin to offer help other than to begin the problem, which you may have already done. I am sure you have tried to work the problem, so show our helpers what your efforts have been. This shows them where you may be going wrong, and how to best help.
 

chisigma

Well-known member
Feb 13, 2012
1,704
If $\displaystyle y=\frac{3}{4}+\frac{3*5}{4*8}+\frac{3*5*7}{4*8*12}+.....\infty$. Then $y^2+2y = $
First we write the series as $\displaystyle y= \sum_{n=1}^{\infty} a_{n}$, where the $a_{n}$ are the solution of the difference equation...

$\displaystyle a_{n+1}= a_{n}\ \frac{2n+3}{4\ (n+1)} = \frac{a_{n}}{2}\ \{1+\frac{1}{2 (n+1)}\},\ a_{0}=1$ (1)

Following the procedure described in...

http://www.mathhelpboards.com/f15/difference-equation-tutorial-draft-part-i-426/

... You find the solution of (1)...

$\displaystyle a_{n}= 2^{1 -n} \prod_{k=1}^{n} (1+\frac{1}{2 k})$ (2)

Now if You 'remember' the formula...

$\displaystyle \prod_{k=1}^{n} (1+\frac{1}{2 k}) = \frac{2}{\sqrt{\pi}}\ \frac{\Gamma(n + \frac{3}{2})}{\Gamma(n+1)}$ (3)

... You arrive to write...

$\displaystyle y = \frac{2}{\sqrt{\pi}}\ \sum_{n=1}^{\infty} \frac{\Gamma(n + \frac{3}{2})}{\Gamma(n+1)}\ 2^{-n}$ (4)

... and 'remembering' also that is...

$\displaystyle \sum_{n=0}^{\infty} \frac{\Gamma(n+\frac{k}{2})}{\Gamma(n+1)}\ x^{n} = \frac{\Gamma(\frac{k}{2})}{(\sqrt{1-x})^{k}}$ (5)

... and...


$\displaystyle \Gamma(\frac{3}{2})= \frac{\sqrt{\pi}}{2}$ (6)

... You obtain finally...

$\displaystyle y=\frac{4}{\sqrt{\pi}} \{\frac{\Gamma(\frac{3}{2})}{(\sqrt{(\frac{1}{2}})^{3}}-\Gamma (\frac{3}{2})\}= 2\ (\sqrt{8}-1)$ (7)

Of course the problem is not trivial and I'm a little surprised that it has been proposed in the Pre-Algebra forum...

Kind regards

$\chi$ $\sigma$
 
Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
If $\displaystyle y=\frac{3}{4}+\frac{3*5}{4*8}+\frac{3*5*7}{4*8*12}+.....\infty$. Then $y^2+2y = $
First hint: $y^2+2y = (y+1)^2 - 1$. So it will be helpful to find $y+1 = 1 + \frac{3}{4}+\frac{3*5}{4*8}+\frac{3*5*7}{4*8*12}+ \ldots.$

Second hint: This looks like a generalised binomial series. In fact, Newton's generalised binomial theorem states that $$(1+x)^s = 1 + \frac s1x + \frac{s(s-1)}{2!}x^2 + \frac{s(s-1)(s-2)}{3!}x^3 + \ldots$$ (and it converges provided that $|x|<1$). Can you force the series for $1+y$ into that form?

Third hint:
$$1+y = 1 + \frac{-\frac32}1\bigl(-\tfrac12\bigr) + \frac{\bigl(-\frac32\bigr) \bigl(-\frac52\bigr)}{2!}\bigl(-\tfrac12\bigr)^2 + \frac{\bigl(-\frac32\bigr) \bigl(-\frac52\bigr) \bigl(-\frac72\bigr)}{3!}\bigl(-\tfrac12\bigr)^3 + \ldots\,.$$
 

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,685
First hint: $y^2+2y = (y+1)^2 - 1$. So it will be helpful to find $y+1 = 1 + \frac{3}{4}+\frac{3*5}{4*8}+\frac{3*5*7}{4*8*12}+ \ldots.$

Second hint: This looks like a generalised binomial series. In fact, Newton's generalised binomial theorem states that $$(1+x)^s = 1 + \frac s1x + \frac{s(s-1)}{2!}x^2 + \frac{s(s-1)(s-2)}{3!}x^3 + \ldots$$ (and it converges provided that $|x|<1$). Can you force the series for $1+y$ into that form?

Third hint:
$$1+y = 1 + \frac{-\frac32}1\bigl(-\tfrac12\bigr) + \frac{\bigl(-\frac32\bigr) \bigl(-\frac52\bigr)}{2!}\bigl(-\tfrac12\bigr)^2 + \frac{\bigl(-\frac32\bigr) \bigl(-\frac52\bigr) \bigl(-\frac72\bigr)}{3!}\bigl(-\tfrac12\bigr)^3 + \ldots\,.$$
I wish I can give as many thanks as possible to you for this solution, Opalg!:p:)(Wink)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I wish I can give as many thanks as possible to you for this solution, Opalg!:p:)(Wink)
Yes, Opalg is certainly one of our best helpers, both in the depth of his knowledge, and also in the lucidity of his posts. We are most fortunate that he is a member here. (Yes)

It is nice when our members take the time to comment on this, and while our helpers do what they do not for accolades, but for the desire to simply be helpful, such comments are certainly appreciated. (Cool)