Find the value of $T=\mathop {\lim }\limits_{n \to \infty } {\left( {1+ \frac{{1+\frac{1}{2}+ \frac{1}{3}+ . +\frac{1}{n}}}{{{n^2}}}} \right)^n}$

Question

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

Answer 1

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 $\ge\frac12$.

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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 $\lim\limits_{n\to\infty}\left(1+\frac{\sum_{k=1}^n\frac1k}{n^2}\right)^n\ge1$, $$ \lim_{n\to\infty}\left(1+\frac{\sum_{k=1}^n\frac1k}{n^2}\right)^n=1\tag4 $$

Answer 2

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 $$\log T \le 0 \Rightarrow \log T = 0 \Rightarrow T=1.$$

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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.

Answer 3

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 $=0.017$%).

Answer 4

You can crudely bound $1+\frac{1}{2} + \cdots + \frac{1}{n}$ by $$\sum_{k=1}^n \frac{1}{k} \le \sum_{k=1}^{\lceil\sqrt{n}\rceil} \frac{1}{k} + \sum_{k=\lceil \sqrt{n} \rceil}^n \frac{1}{k} \le \lceil \sqrt{n}\rceil \cdot 1 + n \cdot \frac{1}{\sqrt{n}} \le 3\sqrt{n},$$ so that when you divide this by $n$, the expression is bounded by $3 \frac{\sqrt{n}}{n} \to 0$.

Answer 5

Your solution is incorrect. The good solution is that $\sum_{k=1}^n \frac1k \sim \log n,$ so that

$$ 1\leq T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + . + \frac{1}{n}}}{{{n^2}}}} \right)^n} \leq \mathop {\lim }\limits_{n \to \infty }\left(1 + \frac{\log n}{n^2} \right)^n$$

For any $k,$ there exists an $N,$ such that $\frac{\log n}{n^2} \leq \frac{1}{k n}$ for $n>N,$ so

$$\mathop {\lim }\limits_{n \to \infty }\left(1 + \frac{\log n}{n^2} \right)^n \leq \exp(\frac1k).$$ It follows that $T=1.$

Answer 6

Your doubt is right - it is not possible to split limit in variable amount of summands, as shows example $1=\lim 1=\lim\left(\frac{1}{n}+\cdots + \frac{1}{n}\right)=\lim \frac{1}{n} +\cdots + \lim\frac{1}{n} = 0$.

On another way you can use representation

$$\sum_{i=1}^{n}\frac{1}{i} = \ln n + \gamma + \varepsilon_{n}= \ln n +O(1), n\rightarrow \infty$$ where $ \varepsilon_{n} \sim \frac{1}{2n}, n\rightarrow \infty$ based on Euler–Mascheroni constant $\gamma$. This also gives you $\frac{{1 + \frac{1}{2} + \frac{1}{3} + . + \frac{1}{n}}}{n} \to 0, n\rightarrow \infty$.

Answer 7

We want to show that $$\displaystyle \lim_{n \to \infty} \dfrac{H_n}{n}=0$$.

We have $\log n = \displaystyle \int_{1}^{n} \dfrac{1}{t} dt$. Also, $H_n = \displaystyle \sum_{r=1}^{n}\dfrac{1}{r} = \dfrac{1}{n}\sum_{r=1}^{n}\dfrac{1}{\frac{r}{n}}$

Thus, by comparison with the Riemann sums of $\log n$, we see that $\log n < H_n < 1 + \log n$.

So we get that, $$\dfrac{\log n}{n} < \dfrac{H_n}{n} < \dfrac{1+\log n}{n}$$

But both the left and right sides of the above inequality tend to $0$ as $n \to \infty$, and hence, by the Sandwich theorem (or Squeeze principle), $$\lim_{n \to \infty} \dfrac{H_n}{n} = 0$$