<s>[INST] Find $\lim \limits_{n \to \infty}\left( 1 + \sqrt{2} + \sqrt[3]{3} + \dots \sqrt[n]{n} \right) \ln{2n+1 \over n}$ I am looking for
$$\lim \limits_{n \to \infty}\left( 1 + \sqrt{2} + \sqrt[3]{3} + \dots \sqrt[n]{n} \right) \ln{2n+1 \over n}$$
We notice $\ln{2n+1 \over n} = \ln\left({1 + {n+1 \over n}}\right)$. We also know that ${x \over 1 + x} \le \ln(1+x)$. From this, we get
$${n+1 \over 2n+1} \le \ln\left({1 + {n+1 \over n}}\right)$$
Then we notice $n \le 1 + \sqrt{2} + \sqrt[3]{3} + \dots \sqrt[n]{n}$, so
$$n{n+1 \over 2n+1} \le \left( 1 + \sqrt{2} + \sqrt[3]{3} + \dots \sqrt[n]{n} \right) \ln{2n+1 \over n}$$
$n{n+1 \over 2n+1} \to + \infty$, so
$$\lim \limits_{n \to \infty}\left( 1 + \sqrt{2} + \sqrt[3]{3} + \dots \sqrt[n]{n} \right) \ln{2n+1 \over n} = + \infty$$
Is this reasoning correct?
 [/INST] Your reasoning is correct. But you could say that all terms in bracket (the first part) are greater than 1, and second part goes to $ln(2)$, which is non zero. So the total limit goes to Infinity.
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