Evaluating $\lim_{x\to\infty}\frac{\sum_{r=1} ^x re^\frac{r}{x}}{x^2}$ without using L-Hospital rule?

Question

I need to evaluate the following limit of the indeterminate form "$\frac{\infty}{\infty}$".

$$\lim_{x\to\infty}\frac{\sum_{r=1} ^x re^\frac{r}{x}}{x^2}$$ $$OR$$ $$\lim_{x\to\infty}\frac{e^\frac{1}{x} + 2e^\frac{2}{x} +3e^\frac{3}{x} +4e^\frac{4}{x} +5e^\frac{5}{x}+\ldots+xe^\frac{x}{x}}{x^2}$$

The answer to this limit is unknown to me as I myself framed this question by chance which I'm unable to solve now. Of course it's possible that this question might already exist for long time(but I have not seen anywhere yet).

Your nice help towards the solution to this problem will be highly appreciable.

NOTE-1: I'm a high school student.
NOTE-2: Use of L'Hospital is not allowed.

UPDATE: Folks are saying that this problem is quite tough for a high school student but to your surprise I’ve found the solution & it’s quite easy and I've written the answer below in answers section. The answer to the problem is 1(one) (answer proof 1, answer proof 2)

Answer 1

The solution to this probel is quite SIMPLE as oppposed to others who are claiming it to be tough for a high school student....

the limit to be evaluated is.....

$$\lim_{x\to∞}\Big(\frac{e^\frac{1}{x}+ 2e^\frac{2}{x}+3e^\frac{3}{x}+4e^\frac{4}{x}+.....+xe^\frac{x}{x}}{x^2}\Big)$$

By subsituting $e^\frac{1}{x}=t$ $$\lim_{x\to∞}\Big(\frac{t+ 2t^{2}+3t^{3}+4t^{4}+.....+xt^{x}}{x^2}\Big)$$ $$\tag*{$...3$}$$ to keep things simple let us first calculatethe value of $ t+ 2t^{2}+3t^{3}+4t^{4}+.....+xt^{x} $

$$S=t+ 2t^{2}+3t^{3}+4t^{4}+.....+xt^{x} $$ $$\tag*{$...1$}$$

dividing both sides by $t$

$$\frac{S}{t}=1+ 2t^{1}+3t^{2}+4t^{3}+.....+xt^{x-1}$$ $$\tag*{$...2$}$$

Subtarcting 1 from 2
$$S\Big(\frac{1-t}{t}\Big)=1+ t^{1}+t^{2}+t^{3}+.....+t^{x-1}-xt^x$$

Using the sum of geometric progression $$S\Big(\frac{1-t}{t}\Big)=\frac{t^x-1}{t-1}-xt^x$$ $$S\Big(\frac{1-t}{t}\Big)=\frac{t^x-1-xt^{x+1}+xt^x}{t-1}$$ $$S=\frac{t^{x+1}-t-xt^{x+2}+xt^{x+1}}{-(t-1)^2}$$

Substituting back $t=e^\frac{1}{x}$ $$S=\frac{-e^{1+\frac{1}{x}}+e^{\frac{1}{x}}+xe^{1+\frac{2}{x}}-xe^{1+\frac{1}{x} }}{(e^\frac{1}{x}-1)^2}$$

Putting S back in the equation 3

$$\lim_{x\to∞}\frac{-e^{1+\frac{1}{x}}+e^{\frac{1}{x}}+xe^{1+\frac{2}{x}}-xe^{1+\frac{1}{x}}}{x^2(e^\frac{1}{x}-1)^2}$$

$$\lim_{x\to∞}\frac{-e^{1+\frac{1}{x}}+e^{\frac{1}{x}}+\frac{xe^{1+\frac{1}{x}}(e^{\frac{1}{x}}-1)x}{x}}{x^2\frac{(e^\frac{1}{x}-1)^2}{x^2}x^2}$$

By applying the standard limit $$\lim_{x\to 0}\frac{a^x-1}{x}=\log_{e} a$$

$$\frac{-e+1+e}{1}$$ $$1$$

$$CHEERS...$$

Answer 2

If you define $$f(x)=xe^x$$ then the sum $$\sum_{k=1}^n\frac1{n^2}ke^{k/n}$$ is $$\sum_{k=1}^n\frac1nf\left(\frac kn\right)$$ which tends to $$\int_0^1f(x)dx=(1-1)e^1-(0-1)e^0=1$$

NOTE: this problem is very difficult for a high school student.

Answer 3

As @ajotatxe already wrote, "this problem is very difficult for a high school student"

Considering $$S_x=\frac{\sum_{r=1} ^x re^\frac{r}{x}}{x^2}$$ the numerator is "just" $$\sum_{r=1} ^x re^\frac{r}{x}=\frac{e^{\frac{1}{x}} \left(e^{\frac{1}{x}+1} x-e x-e+1\right)}{\left(e^{\frac{1}{x}}-1\right)^2}$$ Using Taylor expansion (this is even less trivial than L'Hospital rule) for large values of $x$, you would have $$\sum_{r=1} ^x re^\frac{r}{x}=x^2+\frac{e }{2}x+\frac{2e-1}{12}+O\left(\frac{1}{x^2}\right)$$ making $$S_x=1+\frac{e }{2x}+O\left(\frac{1}{x^2}\right)$$ which shows the limit and also how it is approached.

Just for the fun of it, use your pocket calculator for $x=5$; you will get $$S_5=\frac{1}{25} \left(e^{1/5}+2 e^{2/5}+3 e^{3/5}+4 e^{4/5}+5 e\right)\approx 1.287$$ while the above expansion would give $1+\frac{e}{10}\approx 1.272$