# What is the exact value of the series $\sqrt{1+2\times\sqrt{1+3\times\sqrt{1+4\times\sqrt\cdot…}}}$ [duplicate]

Kindly tell me what is the value of:

$\sqrt{1+2\times\sqrt{1+3\times\sqrt{1+4\times\sqrt\cdot.....}}}$

According to ramanujan ,it is equal to 3

I want to know how...

## marked as duplicate by Hans Lundmark, Alex M., Matthew Conroy, Ross Millikan calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Dec 6 '16 at 22:11

• You should be using backward slashes, \, instead of forward slashes, /, to format your question. – RGS Dec 5 '16 at 11:14
• In third line it is 3,,, not 3!=6,,, – Atul Mishra Dec 5 '16 at 11:14
• What might help : If we replace the square root after the number $n$ with $n+2$ , we get exactly $3$, for example $$\sqrt{1+2\cdot \sqrt{1+3\cdot \sqrt{1+4\cdot 6}}}=3$$ – Peter Dec 5 '16 at 13:51
• I've seen ramanujan's solutions but I am not satisfied that this will work up to infinity....? – Atul Mishra Dec 5 '16 at 14:53

This formula makes no sense. When u write "$\ldots$" it's like saying "and so on" where you expect the reader to understand from context what you are talking about.

For example if commonly when u see $1+2+3+4+\ldots$ you would think about a limit of sequence of partial sums therefore you would said $1+2+3+4+\ldots=\infty$ since that sequence does not converge.

However, speaking of ramanujan, he used different context where he stated that $1+2+3+4+\ldots=-{1 \over 12}$ simply because he meant something different by "$\ldots$"

Returning to your question. Your "$\ldots$" does not point to any scheme. ramanujan wrote something like it's 3 because

$3=\sqrt{9}=\sqrt{1+2\cdot4}=\sqrt{1+2\sqrt{16}}=\sqrt{1+2\sqrt{1+3\cdot5}}=\sqrt{1+2\sqrt{1+3\sqrt{25}}}=\sqrt{1+2\sqrt{1+3\sqrt{1+4\cdot6}}}=\ldots=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{\ldots}}}}$

Unfortunately such argument would work for any positive value.

For example $10=\sqrt{100}=\sqrt{1+2\cdot{99 \over 2}}=\sqrt{1+2\sqrt{1+3\cdot{ 9797 \over 12}}}=\ldots$

the weird fraction will become irrelevant