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

Problem with the Approach in the Question
One must take extra care when taking the limit of the sum of an increasing number of terms, each of which is tending to $0$. For example,
$$
\overbrace{\frac1{n+1}+\frac1{n+2}+\frac1{n+3}+\cdots+\frac1{n+n}}^{\text{$n$ terms, each of which tends to $0$}}
$$
If we sum the limits of each term, we get $0$. However, each of the $n$ terms is $\ge\frac1{2n}$, therefore, the $\liminf\limits_{n\to\infty}$ of the sum is $\ge\frac12$.

Inequalities and the Limit
First, we have the bound
$$
\begin{align}
\sum_{k=1}^n\frac1k
&\le1+\sum_{k=2}^n\int_{k-1}^k\frac{\mathrm{d}x}x\tag{1a}\\
&=1+\int_1^n\frac{\mathrm{d}x}x\tag{1b}\\[6pt]
&=1+\log(n)\tag{1c}
\end{align}
$$
Explanation:
$\text{(1a)}$: $\frac1k\le\int_{k-1}^k\frac{\mathrm{d}x}x$
$\text{(1b)}$: the sum of an integral over disjoint intervals
$\phantom{\text{(1b):}}$ is the integral over the union of those intervals
$\text{(1c)}$: integrate
Furthermore, for $x\gt-1$ and $n\ge1$ and $nx\lt1$,
$$
\begin{align}
(1+x)^n
&\le\left(\frac1{1-x}\right)^n\tag{2a}\\
&\le\frac1{1-nx}\tag{2b}
\end{align}
$$
Explanation:
$\text{(2a)}$: $(1+x)(1-x)=1-x^2\le1$
$\text{(2b)}$: Bernoulli's Inequality applied to the reciprocal
Thus,
$$
\begin{align}
\lim_{n\to\infty}\left(1+\frac{\sum_{k=1}^n\frac1k}{n^2}\right)^n
&\le\lim_{n\to\infty}\left(1+\frac{1+\log(n)}{n^2}\right)^n\tag{3a}\\
&\le\lim_{n\to\infty}\frac1{1-\frac{1+\log(n)}n}\tag{3b}\\[3pt]
&=1\tag{3c}
\end{align}
$$
Explanation:
$\text{(3a)}$: apply $(1)$
$\text{(3b)}$: apply $(2)$
$\text{(3c)}$: $\lim\limits_{n\to\infty}\frac{1+\log(n)}n=0$
Therefore, since $\lim\limits_{n\to\infty}\left(1+\frac{\sum_{k=1}^n\frac1k}{n^2}\right)^n\ge1$,
$$
\lim_{n\to\infty}\left(1+\frac{\sum_{k=1}^n\frac1k}{n^2}\right)^n=1\tag4
$$
