# Infinite series sum

#### jacks

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

#### MarkFL

Staff member
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
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

$\chi$ $\sigma$

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#### Opalg

##### MHB Oldtimer
Staff member
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
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!