What is $\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4+\cdots}}}}$ equal to

Question

I am currently a student in sixth form in the UK, I have been trying to solve a problem that my friend gave me for some time now but haven't gotten any where with it I tried going on Google and searching for an answer but couldn't find anything(mostly likely because I don't know how to phrase the question).

The question my friend asked me was this:

$f(n)=\sqrt{n+f(n+1)}$

Find some formula or some way of calculating the output of this function for any value of $n$

For example $f(1)=\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4+...}}}} \approx 1.7579...$

This all started when our teacher asked us to find the value of $\sqrt{1+\sqrt{1+\sqrt{1+...}}}$

When we got the answer we tried to see if we could generalise it and we got $g(n)=\sqrt{n+g(n)} \ $ and where then able to derive the formula $\frac{1+\sqrt{1+4n}}{2} = g(n) = \sqrt{n+\sqrt{n+...}}$

However neither of us have been able to derive such a formula for $f(n)=\sqrt{n+f(n+1)}$

I asked a question similar to this: https://math.stackexchange.com/questions/3222089/rmaltaecursion-like-sequences-which-are-hard-to-relate-recursively — Luke Collins, Dec 08 '19 at 20:53
https://math.stackexchange.com/questions/201193/sum-and-product-of-infinite-radicals — Angina Seng, Dec 08 '19 at 20:56
So what you're saying is that there is no solution or expression to this problem, a bit disappointing but at least I can now stop thinking about this problem and have some peace of mind. Thanks for your comment! — Rxy, Dec 08 '19 at 21:14
@Rxy Unfortunately, many (arguably most) sequences, series, integrals and so on have no closed-form solution. — Jam, Dec 08 '19 at 22:22
Although not strictly relevant, this might serve as a consolation prize: Evaluating the nested radical $\sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + \cdots}}}$ (The answer in that other case is nice and simple - it's $3$.) — Calum Gilhooley, Dec 08 '19 at 22:44
On the edits to the title: the OP chose to use the $\LaTeX{}$ "\sqrt" command in the body of the question, but a non-$\LaTeX{}$ square root symbol in the title. As this must have been a conscious choice, it seemed appropriate to respect it by making only a minimal change, although I agree that the $\LaTeX{}$ form looks much nicer. — Calum Gilhooley, Dec 10 '19 at 13:42

What is $\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4+\cdots}}}}$ equal to

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