How to fully solve this limit evaluation using integration?

[juantheron]

Evaluation of $$\displaystyle \lim_{n\rightarrow \infty}\sum^{n}_{k=1}\bigg(\frac{k}{n^2}\bigg)^{\frac{k}{n^2}+1}$$

 

[SatyaDas]

Evaluation of $$\displaystyle \lim_{n\rightarrow \infty}\sum^{n}_{k=1}\bigg(\frac{k}{n^2}\bigg)^{\frac{k}{n^2}+1}$$

Spoiler

We can set the followings:
$$
dx=\frac{1}{n}$$
and
$$
x=\frac{k}{n}
$$
As $$n\rightarrow \infty$$ $$\sum$$ is replaced with $$\int$$.
So, we finally have:

$$
\int_{0}^{1} (x dx)^{1 +x dx}
=> \int_{0}^{1} x dx ((x dx)^{x})^{dx}
$$

Term inside () in above integration will be one because in limiting case $$dx \rightarrow 0.$$ To be honest I might be taking a leap here. :)
So, the problem actually reduces to:

$$
\int_{0}^{1} x dx
$$

Whose value is 0.5.

 

[Opalg]

Evaluation of $$\displaystyle \lim_{n\rightarrow \infty}\sum^{n}_{k=1}\bigg(\frac{k}{n^2}\bigg)^{\frac{k}{n^2}+1}$$

Spoiler

We can set the followings:
$$
dx=\frac{1}{n}$$
and
$$
x=\frac{k}{n}
$$
As $$n\rightarrow \infty$$ $$\sum$$ is replaced with $$\int$$.
So, we finally have:

$$
\int_{0}^{1} (x dx)^{1 +x dx}
=> \int_{0}^{1} x dx ((x dx)^{x})^{dx}
$$

Term inside () in above integration will be one because in limiting case $$dx \rightarrow 0.$$ To be honest I might be taking a leap here. :)
So, the problem actually reduces to:

$$
\int_{0}^{1} x dx
$$

Whose value is 0.5.

Brilliant intuition, Satya! But this is a math forum, not an engineering forum, so your argument needs a few sticking plasters to make it rigorous.

[sp]Define a function $f(x)$ for $x\geqslant0$ by $$f(x) = \begin{cases}x^{x+1}&(x>0),\\0&(x=0). \end{cases}$$ Then $f$ is continuously differentiable, with $$f'(x) = \begin{cases}x^x(x + \ln x + 1)&(x>0),\\ 1&(x=0).\end{cases}$$ Strictly speaking, the derivative at $x=0$ is only a one-sided derivative. But that is the crucial fact that is needed, because it implies that $f(x)$ is very close to $x$ when $x$ is small. More precisely, given $\varepsilon>0$ there exists $\delta>0$ such that $(1-\varepsilon)x < f(x) < (1+\varepsilon)x$ whenever $0<x<\delta$.

Now choose $n$ with $\frac1n<\delta$, and let $1\leqslant k\leqslant n$. We can then put $x = \frac k{n^2}$ in the above inequalities to get $$(1-\varepsilon)\frac k{n^2} < \bigg(\frac{k}{n^2}\bigg)^{\frac{k}{n^2}+1} < (1+\varepsilon)\frac k{n^2}.$$ Sum that from $k=1$ to $n$, using the fact that $$\sum_{k=1}^n k = \tfrac12n(n+1)$$, getting $$(1-\varepsilon)\frac{n(n+1)}{2n^2} < \sum_{k=1}^n \bigg(\frac{k}{n^2}\bigg)^{\frac{k}{n^2}+1} < (1 + \varepsilon)\frac{n(n+1)}{2n^2}.$$ Let $n\to\infty$ to get $$\frac12(1-\varepsilon) \leqslant \lim_{n\to\infty} \sum_{k=1}^n \bigg(\frac{k}{n^2}\bigg)^{\frac{k}{n^2}+1} \leqslant \frac12(1+\varepsilon).$$ Finally, let $\varepsilon\to0$ to see that $$\lim_{n\to\infty} \sum_{k=1}^n \bigg(\frac{k}{n^2}\bigg)^{\frac{k}{n^2}+1} = \frac12$$.

(The last part of that argument could alternatively be done by using a Riemann sum approximation to an integral, which is what Satya was doing.)

[/sp]

 

[SatyaDas]

Wow. I had struggled to formalize my solution. You solved this problem in the way it should be solved. Great.

 

[I like Serena]

Thank you Opalg. I was struggling to find a rigorous proof as well.
Using your solution, I could finish mine.
Just as a slight alternative, here is my solution.

Spoiler

Let's take as a given that:
$$\lim_{x\to 0^+} x^x = 1 \tag{1}$$
We can prove it separately (indirectly) with l'Hôpital's rule, but I'll keep that out of scope for now.

It means that for every $\varepsilon >0$ there is an $\delta >0$ such that for every $0<x<\delta$ we have: $1-\varepsilon < x^x < 1 + \varepsilon$.
Moreover, for $N$ sufficiently big we have that $0 < \frac 1N < \delta$.
That is, there is an $N$ such that for all $n> N$ and for all $1\le k \le n$ we have that $0<\frac k{n^2} \le \frac n{n^2} < \frac 1N < \delta$.
And therefore:
$$1-\varepsilon < \left(\frac k{n^2}\right)^{\frac k{n^2}} < 1 + \varepsilon \tag{2}$$

Let $s_n$ be the summation in the problem statement up to $n$. Then:
\begin{aligned}
s_n = \sum^{n}_{k=1}\bigg(\frac{k}{n^2}\bigg)^{\frac{k}{n^2}+1}
&= \left(\frac 1{n^2}\right)^{\frac 1{n^2}+1} + \left(\frac 2{n^2}\right)^{\frac 2{n^2}+1} + \ldots + \left(\frac n{n^2}\right)^{\frac n{n^2}+1} \\
&= \frac 1{n^2}\left[ \left(\frac 1{n^2}\right)^{\frac 1{n^2}} + \left(\frac 2{n^2}\right)^{\frac 2{n^2}}\cdot 2 + \ldots + \left(\frac n{n^2}\right)^{\frac n{n^2}} \cdot n \right]
\end{aligned}
Using (2), we get the following.
For every $\varepsilon >0$ there is an $N$ such that for all $n> N$:
\begin{aligned}\frac 1{n^2}\big[ (1-\varepsilon) + (1-\varepsilon)2 + \ldots + (1-\varepsilon)n \big]
&< s_n < \frac 1{n^2}\big[ (1+\varepsilon) + (1+\varepsilon)2 + \ldots + (1+\varepsilon)n \big] \\
\frac {1-\varepsilon}{n^2}\cdot \frac 12n(n+1) &< s_n < \frac {1+\varepsilon}{n^2}\cdot \frac 12n(n+1) \\
\frac 12 (1-\varepsilon) &< s_n < \frac 12(1+\varepsilon)(1+\frac 1n)
\end{aligned}
Now let $\varepsilon \to 0^+$ and $n\to\infty$ and we get:
$$\lim_{n\to\infty}\sum^{n}_{k=1}\bigg(\frac{k}{n^2}\bigg)^{\frac{k}{n^2}+1} = \lim_{n\to\infty} s_n = \frac 12$$

 

[I like Serena]

One more variation. This time with an integral as Satya suggested and with sticking plasters from Opalg.

Spoiler

Let $s_n = \sum^{n}_{k=1}\left(\frac{k}{n^2}\right)^{\frac{k}{n^2}+1}$.
Let $x_k = \frac kn$ and $\Delta x = \frac 1n$.

Then it follows from the definition of a Riemann integral that:
$$\lim_{n\to\infty}\sum_{k=1}^n x_k\Delta x = \int_0^1 x\,dx \tag 1$$

Using $\lim\limits_{x\to 0^+} x^x = 1$ we can find the following, as explained in my previous solution.

For every $\varepsilon > 0$ there is an $N$ such that for all $n>N$ and all $1\le k \le n$:
$$1-\varepsilon < (x_k \Delta x)^{x_k \Delta x} < 1+\varepsilon \tag 2$$

We have:
$$
s_n =\sum^{n}_{k=1}\bigg(\frac{k}{n^2}\bigg)^{\frac{k}{n^2}+1}
= \sum^{n}_{k=1}(x_k\Delta x)^{x_k\Delta x+1}
= \sum^{n}_{k=1}x_k\Delta x(x_k\Delta x)^{x_k\Delta x}
$$

Therefore, using (2) for $n>N$:
$$
\sum^{n}_{k=1}x_k\Delta x(1-\varepsilon) < s_n < \sum^{n}_{k=1}x_k\Delta x(1+\varepsilon)\\
(1-\varepsilon)\lim_{n\to\infty}\sum^{n}_{k=1}x_k\Delta x \le \lim_{n\to\infty} s_n \le (1+\varepsilon)\lim_{n\to\infty}\sum^{n}_{k=1}x_k\Delta x\\
$$
With (1) we get:
$$(1-\varepsilon) \int_0^1 x\,dx \le \lim_{n\to\infty}s_n \le (1+\varepsilon) \int_0^1 x\,dx$$
which holds true for every $\varepsilon>0$.

Thus:
$$\lim_{n\to\infty} s_n = \int_0^1 x\,dx = \frac 12$$

 

[SatyaDas]

One more variation. This time with an integral as Satya suggested.

Spoiler

Let $s_n = \sum^{n}_{k=1}\left(\frac{k}{n^2}\right)^{\frac{k}{n^2}+1}$.
Let $x_k = \frac kn$ and $\Delta x = \frac 1n$.

Then it follows from the definition of a Riemann integral that:
$$\lim_{n\to\infty}\sum_{k=1}^n x_k\Delta x = \int_0^1 x\,dx \tag 1$$

Using $\lim\limits_{x\to 0^+} x^x = 1$ we can find the following, as explained in my previous solution.

For every $\varepsilon > 0$ there is an $N$ such that for all $n>N$ and all $1\le k \le n$:
$$1-\varepsilon < (x_k \Delta x)^{x_k \Delta x} < 1+\varepsilon \tag 2$$

We have:
$$
s_n =\sum^{n}_{k=1}\bigg(\frac{k}{n^2}\bigg)^{\frac{k}{n^2}+1}
= \sum^{n}_{k=1}(x_k\Delta x)^{x_k\Delta x+1}
= \sum^{n}_{k=1}x_k\Delta x(x_k\Delta x)^{x_k\Delta x}
$$

Therefore, using (2) for $n>N$:
$$
\sum^{n}_{k=1}x_k\Delta x(1-\varepsilon) < s_n < \sum^{n}_{k=1}x_k\Delta x(1+\varepsilon)\\
(1-\varepsilon)\lim_{n\to\infty}\sum^{n}_{k=1}x_k\Delta x \le \lim_{n\to\infty} s_n \le (1+\varepsilon)\lim_{n\to\infty}\sum^{n}_{k=1}x_k\Delta x\\
$$
With (1) we get:
$$(1-\varepsilon) \int_0^1 x\,dx \le \lim_{n\to\infty}s_n \le (1+\varepsilon) \int_0^1 x\,dx$$
which holds true for every $\epsilon>0$.

Thus:
$$\lim_{n\to\infty} s_n = \int_0^1 x\,dx = \frac 12$$

This solution in my view is the simplest and still rigorous.

 

[juantheron]

Thanks friends for yours fantastic solutions

i have solved it using Integration.

after seeing above solutions by opalg and klass van have seems that my solution is partial (Not fully satisfactory)