Limit of a sequence defined by a recurrence

Question

Let $x_1$ and $x_2$ be positive real numbers and define, for $n>2$:$$\displaystyle x_{n+1}=\sum_{k=1}^{n}\sqrt[n]{x_k}$$

Evaluate $\displaystyle \lim_{n\to\infty}\frac{x_n-n}{\ln n}$

Answer 1

I am absolutely not sure the answer I will give is correct, so please pay attention to details!

Let us assume $x_1, x_2 >1$, now $$\lim_{n\to\infty}\frac{x_n-n}{\ln n}=\lim_{n\to\infty}\frac{\sum_{k=1}^{n}\sqrt[n]{x_k}-1}{\ln n}$$Now we can observe that the letter series diverges by ratio test with the harmonic series. In particular we have that $\forall \epsilon ,\exists \nu_\epsilon: \forall n> \nu_\epsilon, (1+\epsilon)\frac{1}{k}>\sqrt[n]{x_k}-1>(1-\epsilon)\frac{1}{k}.$ Let us consider $c=\sum^{\nu_\epsilon}\sqrt[n]{x_k}-1$, now we know that: $$ \frac{c+(1-\epsilon)\sum_{\nu_\epsilon}\frac{1}{k}}{\ln(n)} \leq \frac{\sum_{k=1}^{n}\sqrt[n]{x_k}-1}{\ln (n)} \leq \frac{c+(1+\epsilon)\sum_{\nu_\epsilon}\frac{1}{k}}{\ln(n)} $$ Now we use that $\sum_0^n \frac{1}{k} - ln(n+1)\rightarrow \gamma$, the Euler-Mascheroni constant, which means that $\sum_{\nu_\epsilon}^n \frac{1}{k} - ln(n+1)\rightarrow c'=\gamma-\sum_0^{\nu_\epsilon} \frac{1}{k}$, so we get: $$\frac{c+(1-\epsilon) \left( \sum_{\nu_\epsilon}\frac{1}{k}-\ln(n+1) +\ln(n+1)\right)}{\ln(n)}\leq \frac{\sum_{k=1}^{n}\sqrt[n]{x_k}-1}{\ln (n)} \leq \frac{c +(1+\epsilon) \left( \sum_{\nu_\epsilon}\frac{1}{k}-\ln(n+1) +\ln(n+1)\right)}{\ln(n)},$$ which in terms of limits means: $$1-\epsilon \leq \lim_{n\to\infty}\frac{x_n-n}{\ln n} \leq 1+\epsilon,$$ $\forall \epsilon >0$. For $x_1, x_2 <1$ there should be only some sign swaps when doing the ratio test.