Question

I am trying to solve this interesting question by Ramanujan

Answer 1

Compute the value of :

\[\large{\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+...}}}}}\]

Answer 2

I am thinking of using calculus, but not sure how to ...

Answer 3

Have you solved any of his other ones with square roots before?

Answer 4

Well this was the only one I got in my book. :(

But, I will look for others

Answer 5

They're usually of the form like \[\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+...}}}}\] try solving that, if you haven't already, it's definitely worth your time. As for this one, it appears to be similar.

Answer 6

Reminds me of the golden ratio?

Answer 7

*is the golden ratio.

Answer 8

Well okay first I think I will solve this:

\[\large{x = \sqrt{1+x}}\]

\[\large{\implies x^2 = 1+x}\]

\[\large{\implies x^2 - x - 1 = 0}\]

\[\large{\implies x = \cfrac{1\pm \sqrt{1+4}}{2}}\]

Answer 9

\[\large{x = \cfrac{1\pm\sqrt{5}}{2}}\]

Here, \(\large{x = \sqrt{1+\sqrt{1+...}}}\)

Answer 10

Now that you're done with that, the original one is fairly similar, but you have to be a little more clever when you solve it. The hint I'll give for the way I solved it was to solve a more general case.

Answer 11

Okay , let me think of it a little

Answer 12

I think kanui already gave the solution

Answer 13

Okay I have some general formula for representing a term of this problem:

\[\large{f(n) = n\sqrt{1+f(n+1)}}\]

Answer 14

what is the n outside the radical doing?

Answer 15

It represents the nth sqrt

Answer 16

I am trying to represent a general formula

Answer 17

It needs to be there for each successive term inside. \[f(1)=1\sqrt{1+f(2)}=1\sqrt{1+2\sqrt{f(3)}}\]

Although I didn't solve the problem going this route, but this might work as well.

Answer 18

This series can be generalised by this:

\[\large{n\sqrt{1+(n+1)\sqrt{1+(n+2)\sqrt{1+...}}}}\]

Answer 19

=)

Answer 20

Now, since,

\(\large{(n+2)^2 = n^2 + 4n + 4 = (n+1)(n+3) + 1}\)

So,

\(\large{(n+2) = \sqrt{1+(n+1)(n+3)} = \sqrt{1+(n+1)(n+1+2)}}\)

So,

\[\large{n(n+2) = n\sqrt{1+(n+1)(n+1+2)}}\]

Now, let \[\large{f(n) = n(n+2)}\]

Then,

\[\large{f(n) = n(n+2) = n\sqrt{1+(n+1)(n+1+2)}}\]

\[\large{\implies f(n) = n\sqrt{1+f(n+1)}}\]

Answer 21

By this I get,

\[\large{f(n) = n\sqrt{1+(n+1)\sqrt{1+(n+2)\sqrt{1+...}}}}\]

\[\large{\implies n(n+2) = n\sqrt{1+(n+1)\sqrt{1+(n+2)\sqrt{1+...}}}}\]

Answer 22

Thus, for given series , n = 1. So,

\[\large{\sqrt{1+2\sqrt{1+3\sqrt{1+...}}} = 3}\]

Answer 23

*sigh* it worked. I calculated the answer for n = 1,2,3,4 and so on to 6. And tried to observe some rule and got that n(n+2) part.

Answer 24

Thanks @Kainui it worked :)

Answer 25

That was entertaining, thank you @vishweshshrimali5 :P

Answer 26

:)

Answer 27

oh yeah definitely entertaining

Answer 28

Interesting, I think when I solved it I did it wrong! Hahaha!

\[y=\sqrt{1+n \sqrt{1+(n+1)\sqrt{1+(n+2)\sqrt{...}}}} \] \[\frac{y^2-1}{n}= \sqrt{1+(n+1)\sqrt{1+(n+2)\sqrt{...}}}\] It was then at this point that I made the substitution n+1=k \[\frac{y^2-1}{k-1}= \sqrt{1+k \sqrt{1+(k+1)\sqrt{1+(k+2)\sqrt{...}}}}\] From there I essentially made my mistake and said \[\frac{y^2-1}{k-1}= y\] and plugged in k=2 Hahahaha. Wow.

Answer 29

Even that solution is very interesting. Forget the mistake :)

Answer 30

I don't think this question is done. You have found the right answer, but only by doing a brute force attack on it. I would like to keep trying to find the "right" way to solve this.

Answer 31

Yeah you are right. There is no best solution to any question :)

Answer 32

I only say that because I think we are on the verge of discovering something useful and interesting. I don't think this is how Ramanujan would have solved it is what I mean.

Answer 33

Yeah! He was truly a great mathematician and above all a great man.

Answer 34

oh the mistake was that y and k are different variables ?

Answer 35

*n and k

Answer 36

Yeah, exactly. It's sort of weird. I had just assumed sort of that they were arbitrary, but in reality they really weren't.

Answer 37

Well I would try to find out some more similar general formulae.

Answer 38

I am thinking about this right now... Hmm...

Answer 39

bit its a limit ! it shouldn't matter for the infinite sum or root.. right ?

Answer 40

looking at it again, it looks fine to me.. hmm

Answer 41

Hmm, well I think it does. Another reason why it's wrong is because if you evaluate it like I say, the answer I got was the golden ratio. It seems like this answer HAS to be greater than the golden ratio, since the golden ratio is also posted in this exact same question by me and the only difference is it lacks the increasing coefficients of each successive square root, so in a sense it makes sense since the golden ratio is about 2.7 and the answer here is 3.

Answer 42

ahh i see y changes when we make the substitution n+1 = k

Answer 43

@ganeshie8 I know, it seems so good and I thought it was right until the very end when he solved it... and got a different answer! haha.

Here is an idea, maybe equating it with this will show us something interesting. Probably not. Still thinking.

\[3= \sum_{n=0}^\infty \frac{2}{3}^n\]

Answer 44

tried something like that, kanui

Answer 45

Does it help us to express it the opposite direction using squares? \[f(n)=(\frac{f(n-1)}{n-1})^2-1\]

Answer 46

Depends on what is f(n).. :)

Answer 47

I got this very interesting formula given by Ramanujan :

\[\large{x+n+a = \sqrt{ax + (n+a)^2 +x\sqrt{a(x+n) + (n+a)^2 + (x+n)\sqrt{...}}}}\]

Answer 48

This is a general expression for solving such type of square roots.

Answer 49

We can prove this formula by assuming:

\[\large{f(x,a,n) = (x+a+n)}\]

Answer 50

Since,

\[\large{(x+a+n)^2 = (n+a)^2 + x^2 + 2x(a+n)}\]
\[\large{= (n+a)^2 + ax + x(x+a+2n)}\]
\[\large{= (n+a)^2 + ax + xf(a,x,n+1)}\]

Answer 51

\[\prod_{i=1}^{\infty} n^{1/2^i} \left( \sum_{i=1}^{\infty} [1+(i+1)] \right)\]

Answer 52

is that even valid?

Answer 53

HAHA

Answer 54

Not so sure about its validity @nincompoop . I am trying to check it for different values of n and i.

Answer 55

go ahead
I think the sequence is correct, but I am not sure if the notation is mathematically acceptable

Answer 56

Okay I tried putting n =1.

Answer 57

I got 3*4*....

Answer 58

square root of a square root gives n^(1/4) then root inside of it becomes n^(1/8)
it goes on, hence n^(1/(2^i))

Answer 59

Yeah but the next square root must be multiplied with (i+1) only.

Answer 60

the index part is figured out
so we just need the summation of radicand

Answer 61

In fact, I think its the summation part which has some mistake.

Answer 62

Yeah, I agree with you nin

Answer 63

1+(i+1)

Answer 64

i = 2
1+(2+1) = 3
1+(3+1) = 4
it goes on

Answer 65

your initial radicand is 3 then it goes on by a step

Answer 66

But, that's the problem my initial radicand is not 3.

Answer 67

look at the sequence
1+2
1+3
1+4
1+5

Answer 68

ohhhhhhhhhhhmybad

Answer 69

HAHA I need some sleep

Answer 70

I'll do this after my exam

Answer 71

Well I am very near of figuring something similar to your representation @nincompoop

Answer 72

you can define the product notation as a variable

Answer 73

if that's legal

Answer 74

Well I think I also did something wrong.

Answer 75

I tried to use derive some relation between the number of square roots the expression has above it and the number which is added to n.

Answer 76

I am going to solve some other question for time being. Its postpone this task for now.

Answer 77

* I think its best to postpone this task for now.

Answer 78

I've been busy doing other things, interesting.