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jacks

Well-known member
Apr 5, 2012
226
Evaluation of \(\displaystyle \displaystyle \lim_{n\rightarrow \infty}\sum^{n}_{k=1}\bigg(\frac{k}{n^2}\bigg)^{\frac{k}{n^2}+1}\)
 

Satya

New member
Sep 13, 2019
11
Evaluation of \(\displaystyle \displaystyle \lim_{n\rightarrow \infty}\sum^{n}_{k=1}\bigg(\frac{k}{n^2}\bigg)^{\frac{k}{n^2}+1}\)
We can set the followings:
\(\displaystyle
dx=\frac{1}{n}\)
and
\(\displaystyle
x=\frac{k}{n}
\)
As \(\displaystyle n\rightarrow \infty\) \(\displaystyle \sum\) is replaced with \(\displaystyle \int\).
So, we finally have:

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

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

\(\displaystyle
\int_{0}^{1} x dx
\)

Whose value is 0.5.
 
Last edited by a moderator:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,697
Evaluation of \(\displaystyle \displaystyle \lim_{n\rightarrow \infty}\sum^{n}_{k=1}\bigg(\frac{k}{n^2}\bigg)^{\frac{k}{n^2}+1}\)
We can set the followings:
\(\displaystyle
dx=\frac{1}{n}\)
and
\(\displaystyle
x=\frac{k}{n}
\)
As \(\displaystyle n\rightarrow \infty\) \(\displaystyle \sum\) is replaced with \(\displaystyle \int\).
So, we finally have:

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

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

\(\displaystyle
\int_{0}^{1} x dx
\)

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.

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 \(\displaystyle \sum_{k=1}^n k = \tfrac12n(n+1)\), 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 \(\displaystyle \lim_{n\to\infty} \sum_{k=1}^n \bigg(\frac{k}{n^2}\bigg)^{\frac{k}{n^2}+1} = \frac12\).

(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.)

 
Last edited:

Satya

New member
Sep 13, 2019
11
Wow. I had struggled to formalize my solution. You solved this problem in the way it should be solved. Great.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,736
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.

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:
$$\lim_{n\to\infty}\sum^{n}_{k=1}\bigg(\frac{k}{n^2}\bigg)^{\frac{k}{n^2}+1} = \lim_{n\to\infty} s_n = \frac 12$$
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,736
One more variation. This time with an integral as Satya suggested and with sticking plasters from Opalg .

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

Satya

New member
Sep 13, 2019
11
One more variation. This time with an integral as Satya suggested.

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.
 

jacks

Well-known member
Apr 5, 2012
226
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)