ParthKohli
  • ParthKohli
\[1 + \frac{1}{4} + \frac{1\cdot 3}{4\cdot 8 } + \frac{1\cdot 3 \cdot 5}{4\cdot 8 \cdot 12} + \cdots\]
Mathematics
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
ParthKohli
  • ParthKohli
@ganeshie8
seascorpion1
  • seascorpion1
Your series simplifies to:\[\frac{(2n+1)!!}{(4n)!!}\] for n >=0
ParthKohli
  • ParthKohli
what does it converge to

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More answers

seascorpion1
  • seascorpion1
I'm a bit rusty on this topic but I'll be back in 5mins with an explanation.
DLS
  • DLS
1/2 ?
ParthKohli
  • ParthKohli
nahi re, the first term itself is greater than 1/2.
DLS
  • DLS
That is what @seascorpion1 's simplification converge to
DLS
  • DLS
3/2 ?
ParthKohli
  • ParthKohli
No, the answer is \(\sqrt 2\). You should recognise a Maclaurin series in there.
seascorpion1
  • seascorpion1
Sorry I couldn't help in time but I learned from your answer.
seascorpion1
  • seascorpion1
Are you sure, I get\[\frac{5\sqrt[8]{e}}{4}\]when I used wolfram to check it.
DLS
  • DLS
I guess they are almost the same..accurate upto 2 decimal places
dan815
  • dan815
|dw:1449420019624:dw|
anonymous
  • anonymous
@dan815 \((2n+1)!\neq1\times3\times5\times\cdots\)
dan815
  • dan815
ignore that lol
ganeshie8
  • ganeshie8
\[\sum\limits_{n=0}^{\infty} \dbinom{2n}{n} \dfrac{1}{8^n}\]
ParthKohli
  • ParthKohli
GANESHIE! WHERE HAVE YOU BEEN?
DLS
  • DLS
where are you even now :|
ganeshie8
  • ganeshie8
Heyy sorry watching movie...
ganeshie8
  • ganeshie8
that is maclaurin series of \(1/\sqrt{1-4x}\) see http://math.stackexchange.com/questions/205898/how-to-show-that-1-over-sqrt1-4x-generate-sum-n-0-infty-binom2n
ParthKohli
  • ParthKohli
Great, well done!
DLS
  • DLS
http://ideone.com/ZFPHpC 994!!
DLS
  • DLS
@ganeshie8 meri bhi thodi help pleaz :/
ganeshie8
  • ganeshie8
hey i saw that question on pigeonhole principle a quick google game me lots of links with solutions...
ganeshie8
  • ganeshie8
*gave
ParthKohli
  • ParthKohli
@dls lol do you remember organic?
DLS
  • DLS
yeah..i know its a popular question...but still confused :| just help me with 2-3 questions of this so I can do that in finals :(
DLS
  • DLS
yeah thodi bahut
ganeshie8
  • ganeshie8
Consider the remainders when integers are divided by \(n\). What are the possible values for remainders ?
DLS
  • DLS
0..n-1
ganeshie8
  • ganeshie8
How many are they ?
DLS
  • DLS
n
ganeshie8
  • ganeshie8
Now consider below "n+1" integers : \[1,~11,~111,~\ldots,~(111\ldots \text{n+1 times})\]
DLS
  • DLS
why are we considering 1s ?
ganeshie8
  • ganeshie8
You will see the exact reason in the end
DLS
  • DLS
alright..
ganeshie8
  • ganeshie8
Divide each of them by the integer \(n\) and look at the remainders
ganeshie8
  • ganeshie8
Would you agree that at least two of the "n+1" integers will have the same remainder ?
ganeshie8
  • ganeshie8
you have "n+1" different integers (pigeons) and you know that there are "n" possible remainders (holes)
DLS
  • DLS
all of them are having the same remainder :o if I consider n=5
ganeshie8
  • ganeshie8
By pigeonhole principle one hole contains at least two pigeons, yes ?
DLS
  • DLS
yes..that's true..
DLS
  • DLS
but still didn't get the reason
ganeshie8
  • ganeshie8
Good. Consider those two pigeons that share the same hole
ganeshie8
  • ganeshie8
We're not done with the proof yet... Is everything clear so far ?
DLS
  • DLS
yes..let's say 1 and 1111 with n=6 give the same remainder.
ganeshie8
  • ganeshie8
With pigeonhole principle, we don't care about the specific cases
DLS
  • DLS
ah yeah..just for self understanding. :P
ganeshie8
  • ganeshie8
"n+1" pigeons "n" holes By pigeonhole principle, one hole must contain at least two pigeons
ganeshie8
  • ganeshie8
above reasoning is all that is needed
ganeshie8
  • ganeshie8
that is okay :) lets keep going..
DLS
  • DLS
what if we consider the set of numbers in the beginning as 2,22,222..?
ganeshie8
  • ganeshie8
Hold on that thought for a bit, you will see what happens in the end...
DLS
  • DLS
i thought the proof is over :O
ganeshie8
  • ganeshie8
Nope, we're only half done
DLS
  • DLS
alright..
ganeshie8
  • ganeshie8
But may I know why you think the proof is over ?
ganeshie8
  • ganeshie8
Have we proved anything about the existence of a multiple of "n" with all 1's and 0's ?
DLS
  • DLS
I thought the line " By pigeonhole principle, one hole must contain at least two pigeons" marks the end of the proof :P nvm by bad. We haven't proven the real thing yet.
ParthKohli
  • ParthKohli
Do we have to apply Pigeonhole Principle again for that?
ganeshie8
  • ganeshie8
Nope, rest is easy... I'm sure you can finish it off on your own... but let me complete it anyways...
ganeshie8
  • ganeshie8
Consider those two pigeons that share the same hole
ganeshie8
  • ganeshie8
Just to remind you again : In our specific case, the two pigeons are the two integers with all 1's hole is remainder
DLS
  • DLS
Right!
ganeshie8
  • ganeshie8
Say the integers are \(nq+r\) and \(nl+r\)
ganeshie8
  • ganeshie8
subtract the smaller integer form the larger one, what do you get ?
ParthKohli
  • ParthKohli
Beauty!
DLS
  • DLS
something nice with 1s and 0s :P
DLS
  • DLS
|n(q-l)|
ganeshie8
  • ganeshie8
Also it is a multiple of "n" as desired...
ganeshie8
  • ganeshie8
Getting back to your earlier question, what happens if our set has integers with all 2's instead of all 1's ?
DLS
  • DLS
I see..that is why we considered 11111111 :O nice!!
DLS
  • DLS
then we would get 2020200
ganeshie8
  • ganeshie8
It seems we can show that given any integer "\(n\)", there exists a multiple with all \(0\)'s and \(a\)'s
ganeshie8
  • ganeshie8
\(a\) is some fixed digit
ganeshie8
  • ganeshie8
like \(a00a000aa00\)
DLS
  • DLS
yep..by considering aaaaa
DLS
  • DLS
but I couldn't have thought of this from scratch :/
ganeshie8
  • ganeshie8
me neither, i googled
ganeshie8
  • ganeshie8
that is okay, now we can solve similar problems w/o google right ?
DLS
  • DLS
Don't know why it was given as the first example in my textbook LOL anyways..another tiny question..a bit of different kind..I found that on google too but the explanation went over me :/ and yes..I can try atleast :)
ikram002p
  • ikram002p
well i was thinking about binary and converting to bases and then i had this thought 10^(phi 9n)=1 mod 9n for gcd(10,9n)=1 and then wella
ikram002p
  • ikram002p
eh i don't think its proper a=idea anyway xD
ganeshie8
  • ganeshie8
That actually proves that there exists an integer with all 1's that is a multiple of any given integer "n"
ganeshie8
  • ganeshie8
I think I have asked this question a few days back...
ganeshie8
  • ganeshie8
\(10^q-1 = 999\ldots (\text{q times}) =9*111\ldots(\text{q times})\)
ikram002p
  • ikram002p
oh i haven't see =).

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