If $\displaystyle a_{n}=\bigg(\frac{n!}{1\cdot 3 \cdot 5 \cdot 7\cdot...\cdot (2n+1)}\bigg)^2.$

Then $\displaystyle \lim_{n\rightarrow \infty}\bigg(a_{1}+a_{2}+...+a_{n}\bigg)$ is


$(a)$ Does not exists

$(b)$ Greater than $\displaystyle \frac{4}{27}$

$(c)$ Less than $\displaystyle \frac{4}{27}$

$(d)$ None of these

My Try: $$a_{n}=\bigg[\frac{n!\cdot 2\cdot 4 \cdot 6 \cdots (2n)}{1\cdot 2 \cdot 3\cdot 4\cdot \cdots (2n)\cdots (2n+1)}\bigg]^2$$

$$a_{n}=\bigg[\frac{n!\cdot 2 \cdot 4 \cdot 6 \cdots (2n)}{(2n+1)!}\bigg]^2$$

Could some help me to solve it , Thanks

  • 3
    $\begingroup$ $$ \frac{n!}{1\cdot 3 \cdot 5 \cdot 7\cdot\cdot (2n+1)} = \prod_{k=0}^n \frac{k}{2k+1} $$ $\endgroup$ – gt6989b Apr 17 at 14:47

Why, why bounty when @gt6989b gave you an awesome hint :) ? $$a_n=\left(\left(\frac{1}{3}\right)\cdot\left(\frac{2}{5}\right)\cdot...\cdot\left(\frac{n}{2n+1}\right)\right)^2 < \frac{1}{2^{2n}}=\frac{1}{4^n}$$ and (calculating the first 3 terms and using infinite geometric progression) $$\sum\limits_{n=1}a_n = \frac{1457}{11025}+\sum\limits_{n=4} a_n < \frac{1457}{11025}+\sum\limits_{n=4}\frac{1}{4^{n}}=\frac{1457}{11025}+\frac{1}{4^4}\cdot\frac{1}{1-\frac{1}{4}}=\\ \frac{1457}{11025}+\frac{1}{3\cdot 4^3}=\frac{96923}{705600}<\frac{4}{27}$$

  • 1
    $\begingroup$ I find the start of your answer rude. Since a bounty was offered, we can infer that DXT couldn't figure out how to finish. $\endgroup$ – mathworker21 Apr 20 at 5:35
  • $\begingroup$ There is nothing rude in it. It's just a comment to share my opinion that 100 bounty is too much ... with a smile. $\endgroup$ – rtybase Apr 20 at 7:44
  • $\begingroup$ I don't think it's the value of 100 that you're objecting to, since you only mentioned "bounty" in your answer. So I'm confused. If DXT couldn't figure out how to finish, why is it bad for him/her to offer a bounty? $\endgroup$ – mathworker21 Apr 20 at 7:57
  • $\begingroup$ Well, first of all ask yourself why this question has 3 closing votes? Because it contains enough ingredients of a typical question being closed; missing context, providing none or very few and insignificant of own attempts. Then came the bounty offer, nearly within 24h after asking the question, my guess is to stop the question from being closed. And zero updates even after having a very good hint ... My opinion - bounty was not needed. Just pen an paper and a few minutes of work. And if that wasn't enough, refine the question with very specific details. $\endgroup$ – rtybase Apr 20 at 8:15

A standard trick is to look at the function $$f(x)=\sum \left(\frac{n!}{1.3...(2n+1)}\right)^2x^{2n+1}$$ Find a diiferential equation that $f(x)$ satisfies, solve it and find $f(1)$.

  • $\begingroup$ Indeed $\endgroup$ – Norbert Apr 17 at 15:04
  • 2
    $\begingroup$ I've been trying to do this, but I'm not getting anywhere. Are you making a general statement, or have you actually succeeded in this case? $\endgroup$ – saulspatz Apr 17 at 15:10
  • $\begingroup$ The DE would be $4(xf')'=4+x(x(xf)')'$ but no, I was just making a general statement. I haven't thought how to solve it. $\endgroup$ – Empy2 Apr 17 at 16:35
  • $\begingroup$ Just in case it wasn't clear, I guess that Norbert was just sarcastic. $\endgroup$ – Alex M. Apr 19 at 20:15

Using the Wallis formula $$\dfrac\pi2 = \dfrac{2\cdot2}{1\cdot3}\cdot\dfrac{4\cdot4}{3\cdot5}\cdot\dfrac{6\cdot6}{5\cdot7}\dots$$ in the form of $$a_n=\left(\dfrac{1\cdot2\cdot3\dots n}{3\cdot 5\cdot7\dots(2n+1)}\right)^2 < \frac\pi{(2n+1)2^{2n+1}},$$ one can get $$a_1+a_2+a_3+a_4+\dots < \frac19+\dfrac\pi{2^5}\left(\dfrac15+\dfrac1{7\cdot4}+\dfrac1{9\cdot4^2}+\dots\right)$$ $$ < \frac19+\dfrac\pi{5\cdot2^5}\cdot\dfrac1{1-\dfrac14} = \dfrac19+\dfrac\pi{120}\color{brown}{\mathbf{<\dfrac4{27}}}.$$


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