Nested Radicals: $\sqrt{a+\sqrt{2a+\sqrt{3a+\ldots}}}$

In general, this is not known (as far as I know, at least). This is an extension of the so-called Kasner Number, and there is a good paper on the problems of finding closed forms of the Number can be found here (the paper also discusses infinite nested radicals).

This sequence does converge, at least. It's monotone increasing and bounded, so that's handy.

As an interesting note, many people are familiar with the work of Kasner without knowing it. It was Kasner who gave the name "googol" to the number 1 followed by 100 zeroes. Cool.

For the case $a=1$, see OEIS sequence A072449.

Define $a_n$ to be the sequence presented. Then we have the following (for $n$ large enough such that $an>1$): $$ a_n < \sqrt{1+\sqrt{2+..+\sqrt{(n-1)a+na}}}\stackrel{not}=b_n$$

$$b_{n+1}=\sqrt{1+\sqrt{2+...+\sqrt{(n-1)a+\sqrt{na+(n+1)a}}}} $$

For $n$ large enough we have that $b_{n+1}\leq b_n$ since $na\geq \sqrt{(2n+1)a}$ for $n \geq N_0$. Since $a_n$ is bounded above by a decreasing sequence, and $(a_n)$ is clearly increasing, it is convergent. However, I think its limit does not have a closed form. Even the case $a=1$ presented in the other answer does not have a closed form.

Define the sequence $a_n$ recursively by:

$a_1$:=$\sqrt{a}$

$a_n$= $\sqrt{a+a_{n-1}}$

This sequence is monotone non-decreasing, and bounded so the limit exists, and it is equal to y.

Now use the fact that $\sqrt{x}$ is continuous for $x>0$, so that it is sequentially-continuous;

then, lim

$\ \lim_{x\to\infty} \sqrt{a + \sqrt{a_{n-1}}} $=

$\sqrt{\ \lim_{x\to\infty}(a+\sqrt{a_{n-1}}) }$=

$\sqrt{a+\sqrt{\ \lim_{x\to\infty}(a_{n-1})}}$

By convergence, $a_{n-1}\rightarrow$y