progressions

here is something within the syllabus

just recalled an well-known inequality 0\leq \frac{1}{2}.\frac{3}{4}.\frac{5}{6}\cdots\frac{2n-1}{2n}\leq\frac{1}{\sqrt{3n+1}} (we can prove this by induction)

using above

0\leq \frac{1}{2}.\frac{3}{4}.\frac{5}{6}\cdots\frac{2n-1}{2n}.\frac{2n+1}{2n+2}\leq \frac{1}{\sqrt{3n+4}}

taking \lim_{n\rightarrow\infty} we get what we want..

but sir i cudnt have given a better answer keeping the fact in mind dat i havent learnt limits and asymptotic of central binomial coefficient nad stuffs like these.....but i think sir as it is a progressions sum there shud be a simpler way to solve the sum since no one learns limits before progressions....

sir,i know that the logic is not formal but what's wrong in my logic.....
as denomiator>>>numerator the fraction is tending to zero.....

here i finish the proof:( proof within syllabus should/may exist and i invite everyone to think of it,cz i am not doing the same right now )

I am going to use asymptotic of central binomial coefficient. We claim

\lim_{n\rightarrow\infty}\binom{2n}{n} = \sqrt{\frac{2}{\pi}}\frac{4^n}{\sqrt{2n+1}}

Here is an elementary proof.http://planetmath.org/encyclopedia/AsymptoticsOfCentralBinomialCoefficient.html This can be obtained by stirling's formula as well.

Follow 1st and 2nd line in my previous post (post 7).

The given expression , using above, becomes \lim_{n\rightarrow\infty}\frac{2\sqrt{2}}{\sqrt{\pi(2n+1)}} = 0

just learnt this

may be :

given thing

= \lim_{n\rightarrow\infty}\frac{(2n+1)\binom{2n}{n}}{(n+1)4^n}

= \lim_{n\rightarrow\infty}\frac{2\binom{2n}{n}}{4^n}

=\lim_{n\rightarrow\infty}\frac{2\Gamma (n+\frac{1}{2})}{\sqrt{\pi}\Gamma (n+1)}

Isn't it acceptable that --

\lim_{n\rightarrow\infty}\frac{\Gamma (n+\frac{1}{2})}{\Gamma (n+1)}=0 ??

sir i think it is zero bcoz we find dat the denominator is always greater by 1 of each corresponding term in the numerator...so as the no terms is same as the n is very huge as it is tending to infinity,thus the denominator will be quite large as compared to the numerator....so the fraction will tend to zero.....

it is like:
2[12(1-34)+1.32.4(1-56)+.......]
Breaking the brackets..
2[1.12-1.32.4+1.32.4........]
all the rest terms cancel out nd the last term will be 1.3.5........(2n-1)(2n+1)2.4.6.8.......2n(2n+2)
So S_n=2[12-1.3.5........(2n-1)(2n+1)2.4.6.8.......2n(2n+2)]
If n= ∞
1.3.5........(2n-1)(2n+1)2.4.6.8.......2n(2n+2) will be 0...
so S_n will be 1.....

Ya but how to find the sum of the series?