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Vinayak asked for the second expression - which diverges. The first expression is also very interesting, I haven't found the solution to that yet, just estimated the result to be around 4.14031456214126 using python (look at the comments under my solution). If you find the exact result for the first expression, please let me know!

I think that the expression does not converge at all. To show that, we will proof
$n\sqrt{1 + n^2\sqrt{1 + n^4\sqrt{...}}} > n^k$
for all natural numbers $n$ and $k$. Here is the proof by induction on $k$:

START: k = 1
The values of all square roots in the expression are greater than 1, so we have:
$n\sqrt{1 + ... } > n = n^1$

STEP from k to k +1:

We assume $n\sqrt{1 + n^2\sqrt{1 + n^4\sqrt{...}}} > 2^k$ for all $n$ and a particular $k$.

This also applies to $n^2$, so we have:
$n^2\sqrt{1 + n^4\sqrt{...}} > (n^2)^k = n^{2k}$

We conclude:
$n\sqrt{1 + n^2\sqrt{1 + n^4\sqrt{...}}} > n\sqrt{1 + n^{2k}} > n\sqrt{n^{2k}} = n \times n^k = n^{k+1}$

The start of induction and the step of induction show the assumption for all $k$.

We can apply this statement for $n = 2$:

$2\sqrt{1 + 4\sqrt{1 + 16\sqrt{...}}} > 2^k$

This will exceed any real number for large $k$ and does not converge. Therefore, the whole expression does not converge, the value of $x$ is undefined.

Thanks a lot @Finnley Paolella! Just asking of curiosity, do you know of any expressions like this(I mean continuing infinitely0), which look like they won't converge, but they do?

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## Comments

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TopNewest$\sqrt{1+2\sqrt{1+4\sqrt{1+8\sqrt{1+16\sqrt{\cdots}}}}}$ or $\sqrt{1+2\sqrt{1+4\sqrt{1+16\sqrt{1+256\sqrt{\cdots}}}}}$?

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Vinayak asked for the second expression - which diverges. The first expression is also very interesting, I haven't found the solution to that yet, just estimated the result to be around 4.14031456214126 using python (look at the comments under my solution). If you find the exact result for the first expression, please let me know!

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Ok. So I will start with the first.

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I think that the expression does not converge at all. To show that, we will proof $n\sqrt{1 + n^2\sqrt{1 + n^4\sqrt{...}}} > n^k$ for all natural numbers $n$ and $k$. Here is the proof by induction on $k$:

START: k = 1 The values of all square roots in the expression are greater than 1, so we have: $n\sqrt{1 + ... } > n = n^1$

STEP from k to k +1:

We assume $n\sqrt{1 + n^2\sqrt{1 + n^4\sqrt{...}}} > 2^k$ for all $n$ and a particular $k$.

This also applies to $n^2$, so we have: $n^2\sqrt{1 + n^4\sqrt{...}} > (n^2)^k = n^{2k}$

We conclude: $n\sqrt{1 + n^2\sqrt{1 + n^4\sqrt{...}}} > n\sqrt{1 + n^{2k}} > n\sqrt{n^{2k}} = n \times n^k = n^{k+1}$

The start of induction and the step of induction show the assumption for all $k$.

We can apply this statement for $n = 2$:

$2\sqrt{1 + 4\sqrt{1 + 16\sqrt{...}}} > 2^k$

This will exceed any real number for large $k$ and does not converge. Therefore, the whole expression does not converge, the value of $x$ is undefined.

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Thanks a lot @Finnley Paolella! Just asking of curiosity, do you know of any expressions like this(I mean continuing infinitely0), which look like they won't converge, but they do?

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I think the expression $\sqrt{1 + 2\sqrt{1 + 4\sqrt{1 + 8\sqrt{1 + 16\sqrt{1 + \dots}}}}}$ converges.

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@David Vreken, @Mahdi Raza, @Finnley Paolella

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