sum the following to n terms and infinity...
14+1.34.6+1.3.54.6.8 ......
please dont use binomial.....this is a pure progressions sum....
21here 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..
341Here's one way at arriving at an upper bound for the expression (as shubhodip had already indicated):
http://www.goiit.com/posts/list/algebra-very-easyyyyy-75119.htm
1057sir yur statement proves the expression is > than 1n but as n→∞ 1n→0....so it proves that the expression is greater than 0.....
1057but 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....
62no it is not ketan.
both numerator and denominator tend to infinity.
but how are we sure that they relation is >>
and >> is a very relative term. it could mean a ratio of 1:10^100 and still not be equal to 0
"maths bhavna nahi jaanta" :P
1057sir,i know that the logic is not formal but what's wrong in my logic.....
as denomiator>>>numerator the fraction is tending to zero.....
21here 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
62Eureka?
12=12>
34>23
56>34
78>45
and so on
multiply both sides
u have LHS > 1/n :D
proved
62ketan.. that logic will not work!
I was wondering about a formal proof in the class but i could nto thik of one based on JEE syllabus.
infact now that i think even ratio test fails!
@subhodip! no clue buddy!
21may 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 ??
1057sir 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.....
62A question which i have not been able to prove without root test!!
how are we srue that the thing u typed tends to zero?
1057it 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 Sn=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 Sn will be 1.....
