Compute $\lim_{n\to\infty}$ $1+\sqrt{2+(3+(4+\dots+(n)^{\frac{1}{n}})^{\frac{1}{4}})^{\frac{1}{3}}}$

Question

$\lim_{n\to\infty}$ $1+\sqrt{2+(3+(4+\dots+(n)^{\frac{1}{n}})^{\frac{1}{4}})^{\frac{1}{3}}}$

I am trying to use $e^{\lim_{n\to\infty}1+\sqrt{2+(3+(4+\dots+(n)^{\frac{1}{n}})^{\frac{1}{4}})^{\frac{1}{3}}}}$=$\lim_{n\to\infty}e(e^\sqrt{2+(3+(4+\dots+(n)^{\frac{1}{n}})^{\frac{1}{4}})^{\frac{1}{3}}})$to do this problem, but it doesn't make sense.

Also I try to use $\ln(\lim_{n\to\infty}1+\sqrt{2+(3+(4+\dots+(n)^{\frac{1}{n}})^{\frac{1}{4}})^{\frac{1}{3}}})$ it also make no sense...

so how to solve it? And does it converge?

Answer 1

It converges by Herschfeld's convergence theorem${}^{\color{blue}{[1]}}$

Given $0 < r_i \le 1, i = 1,\ldots$ such that the series $S = \sum_{i=1}^\infty r_1\ldots r_i$ converges.
For any $a_1, a_2, \ldots \ge 0$, the necessary and sufficient condition for the sequence $$u_n = (a_1 + (a_2 + \cdots + a_n^{r_n}))^{r_2})^{r_1}$$ to converge is $$\limsup_{n\to\infty} a_n^{r_1\ldots r_n} < +\infty$$

For the nested radical at hand, we have $a_n = n$ and $r_n = \frac{1}{n}$. Since $$\limsup_{n\to\infty} a_n^{r_1\ldots r_n} = \limsup_{n\to\infty} n^{1/n!} = 1$$ The corresponding sequence of nested radical $u_n$ converges.

Notes/References

• $\color{blue}{[1]}$ - Theorem III, Herschfeld, A. On Infinite Radicals. Amer. Math. Monthly 42, 419-429, 1935.