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juantheron

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- #1

juantheron

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- #2

SatyaDas

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\(\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.

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- #3

Opalg

Gold Member

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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.

\(\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.

[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 \(\displaystyle \sum_{k=1}^n k = \tfrac12n(n+1)\), 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.)

[/sp]

Last edited:

- #4

SatyaDas

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- #5

I like Serena

Homework Helper

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Using your solution, I could finish mine.

Just as a slight alternative, here is my solution.

$$\lim_{x\to 0^+} x^x = 1 \tag{1}$$

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:

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

- #6

I like Serena

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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:

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

- #7

SatyaDas

- 22

- 0

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.

- #8

juantheron

- 247

- 1

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)

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