<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] Consider the partial term
$$T_n=\left(1+\frac{H_n}{n^2}\right)^n\implies \log(T_n)=n \log\left(1+\frac{H_n}{n^2}\right)$$
For large $n$
$$H_n=\log (n)+\gamma +\frac{1}{2 n}-\frac{1}{12 n^2}+\frac{1}{120
   n^4}+O\left(\frac{1}{n^6}\right)$$
$$\log\left(1+\frac{H_n}{n^2}\right)=\frac{\log (n)+\gamma }{n^2}+\frac{1}{2 n^3}+O\left(\frac{1}{n^4}\right)$$
$$\log(T_n)=\frac{\log (n)+\gamma }{n}+\frac{1}{2 n^2}+O\left(\frac{1}{n^3}\right)$$
$$T_n=e^{\log(T_n)}=1+\frac{\log (n)+\gamma }{n}+\frac{(\log (n)+\gamma )^2+1}{2
   n^2}+O\left(\frac{1}{n^3}\right)$$ which shows the limit and also how it is approached.
Moreover, this give a shortcut method for the evaluation of $T_n$. Suppose $n=10$. The exact value is
$$T_{10}=\left(\frac{259381}{252000}\right)^{10} \sim 1.334677$$ while the above truncated series gives $\sim 1.334446$ (relative error $=0.017$%).
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