<s>[INST] Find the value of $T=\mathop {\lim }\limits_{n \to \infty } {\left( {1+ \frac{{1+\frac{1}{2}+ \frac{1}{3}+ . +\frac{1}{n}}}{{{n^2}}}} \right)^n}$ I am trying to evaluate
$$T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}}} \right)^n}.$$
My solution is as follow
$$T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}}} \right)^n} \Rightarrow T = {e^{\mathop {\lim }\limits_{n \to \infty } n\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}} - 1} \right)}} = {e^{\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{n}} \right)}}$$
$$T = {e^{\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{2n}} + \frac{1}{{3n}} + \cdots + \frac{1}{{{n^2}}}} \right)}} = {e^{\left( {0 + 0 + \cdots + 0} \right)}} = {e^0} = 1$$
The solution is correct  but I presume my approach $\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdot + \frac{1}{n}}}{n}} \right) \Rightarrow \mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{2n}} + \frac{1}{{3n}} + \cdot + \frac{1}{{{n^2}}}} \right) = 0$ is wrong.
Is there any generalized method
 [/INST] If
$$T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + . + \frac{1}{n}}}{{{n^2}}}} \right)^n}$$
then
$$\log T = \mathop {\lim }\limits_{n \to \infty } n\log{\left( {1 + \frac{H_n}{{{n^2}}}} \right)}\le \mathop {\lim }\limits_{n \to \infty } \frac{H_n}{n}.$$
It can be easily shown that this limit tends to $0$ (for example by Cesaro-Stolz.) So
$$\log T \le 0 \Rightarrow \log T = 0 \Rightarrow T=1.$$

Addendum: Showing that $\mathop {\lim }\limits_{n \to \infty } \frac{H_n}{n}\to0$.
Cesaro-Stolz:
We may set $a_n=H_n$ and $b_n={n}$. (We can apply Cesaro-Stolz since $(n)_{n\ge 1}$ is strictly monotone and divergent. The limit
$$\lim_{n\to\infty}\frac{H_{n+1}-H_n}{n+1-n}=\lim_{n\to\infty}\frac{1}{n+1},$$
exists so
$$\lim_{n\to\infty}\frac{H_n}{n}=\lim_{n\to\infty}\frac{1}{n}=0.$$
Bounding $H_n$
In this question a user proved $H_n<\log(n)+1.$ That means that
$$\lim_{n\to\infty}\frac{H_n}{n}\le\lim_{n\to\infty}\frac{\log(n)+1}{n}=0.$$
Since our oringal limit is non-negative we can conclude $\lim_{n\to\infty}\frac{H_n}{n}=0$.
l'hopital
One may also stick to the classic l'hopital. Differentiating numerator and denominator yields
$$\mathop {\lim }\limits_{n \to \infty } \frac{H_n}{n}=\mathop {\lim }\limits_{n \to \infty } \frac{\pi^2/6-H_n^{(2)}}{1}.$$
Since $H_n^{(2)}\to\pi^2/6$ for $n\to\infty$, the overall limit equals zero too.
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