Ramanujan Square Root

By Andre Rzym on Wednesday, March 20, 2002 - 04:18 pm:

There is an old (to be archived) thread that postulates:

æ
ú
Ö

1+1×

æ
Ö

1+2×

1+3×Ö{¼}

There appeared to be no conclusive proof.

Would anyone care to offer a proof of the limit (including convergence)? I think I have both but would be interested in how others approach it.

Andre

By Ronald Mccuiston on Thursday, March 21, 2002 - 08:54 pm:

I am interested in a proof of a similar expression in David Wells' 'The Penguin Dictionary of Curious and Interesting Numbers.'
Wells gives a reference to Ramanujan: MM v59 23.
The infinte nested root in Wells has a value of 3.
Is the Wells expression correct?
Look at http://mathworld.wolfram.com/NestedRadical.html for a similar infinte nested root.

By Ronald Mccuiston on Tuesday, April 02, 2002 - 08:44 pm:

Andre,
Yes, I want to see your proof.
I am not able to offer a proof of my own.
Ron

By Andre Rzym on Thursday, April 04, 2002 - 09:03 am:

I am going to have to split the proof into parts due to time constraints.

Firstly, as you point out above, the series appears to converge quite rapidly to 2. Which raises the question: If we consider, say the 5th approximation to the series to be a function of x, i.e. define

for what value of x does F(x)=2? A bit of mental arithmetic gives x=49. Doing the same for the 6th approximation gives x=64.

It is no accident that both are perfect squares. The reason is the identity:

k²=1+(k-1)(k+1)=1+(k-1)

_____
Ö(k+1)²

... (1)

Starting with

2=Ö4

we apply (1) with k=2

2=
Ö

1+1×Ö9

apply (1) again with k=3

2=   æ
ú
Ö

1+1×   æ
Ö

1+2×   __
Ö16

and so on.

What we have shown so far is that elements in the sequence
Ö4

Ö

1+1×Ö9

  æ
ú
Ö

1+1×   æ
Ö

1+2×   __
Ö16

are identically equal to 2. What we need to show is that the sequence
Ö1

Ö

1+1×Ö1

æ
Ö

1+1×
Ö

1+2×Ö1

converges to the same thing. I'll do that shortly (unless someone would like to suggest a proof in the meantime).

Andre

By Ronald Mccuiston on Sunday, April 07, 2002 - 04:18 am:

Andre,
Thanks. Please continue.
Ron

By Andre Rzym on Tuesday, April 09, 2002 - 07:49 am:

Here is the convergence argument. Sorry for the delay - it was partly due to time constraints but mostly because I realised that my original convergence argument was wrong! So here is version 2:

Define

S_k=

1+¼+kÖx

Then

_k+1

Now define

D_k=S_k((k+2)²)-S_k(1)=2-S_k(1)

Colloquially, our objective is to prove that D_k® 0 as k®¥

Consider

D_2k=2-S_k(y)

where
y=1+(k+1) æ
Ö

1+(k+2)
Ö

¼

so y > (k^{2^k-1})/(k^{2^k-1})

D_2k < 2-S_k((k^{2^k-1})/k^{2^k-1}))

D_2k < D_k×((k+2)²-(k^{2^k-1})/(k^{2^k-1}))/((k+2)²-1)

As k®¥, we do not even have to use the fact that D_k is diminishing, instead noting that the fraction tends towards zero.
Andre