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I want to show that $\lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{k=1}^n\log\frac{k}{n}=-1$.

Now I could say that $\lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{k=1}^n\log\frac{k}{n}=\int_0^1\log xdx$ but I can't as $\log x$ isn't uniform continuous on the given interval. I found a way which goes over the convergence radius of a power series connected to the term in the limit, but I think one could just make the argument with the riemann sum work somehow?!

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  • $\begingroup$ I do not know if it is a way to use Riemann sums, but note that your sum is $S_n=(\log n!-n\log n)/n$, so Stirling's formula can be a way to the solution.... $\endgroup$ – Kelenner Aug 7 '16 at 12:10
  • $\begingroup$ Thanks, but I really want to know if the 'riemann sum way' would work somehow, as the integral has the same value! $\endgroup$ – user359874 Aug 7 '16 at 13:43
  • $\begingroup$ Why is this 'almost' Riemann sum; isn't it usually solved using this very method? $\endgroup$ – StubbornAtom Aug 7 '16 at 14:23
  • $\begingroup$ @StubbornAtom What do you mean? I thought you can only go over to the integral in case the function is uniform continuous? $\endgroup$ – user359874 Aug 7 '16 at 17:06
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Hint: Draw the graph of $\log x$ and see that the sum $\sum_{k=1}^n \frac{1}{n}\log(k/n)$ represents the sum of the area of boxes "under" the graph of $\log x$. Then try to estimate the error between the sum and $\int_{1/n}^1 \log x dx$ by finding another sum that represents the area of boxes that cover the graph of $\log x$ from $1/n$ to $1$. E.g. for the interval $[(n-1)/n, 1]$ do not use a box of height $0$ (as is used in your sum), but use a box of height $\log((n-1)/n)$.

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Claim: If $f$ is monotone on $(0,1],$ and the improper integral $\int_0^1 f$ converges, then

$$\int_0^1 f = \lim_{n\to \infty} \sum_{k=1}^{n}f(k/n)(1/n).$$

Proof: WLOG $f\ge 0$ and $f$ is decreasing. Then

$$\int_{1/n}^1 f \le \sum_{k=1}^{n}f(k/n)(1/n) \le \int_0^1 f.$$

This follows by staring down areas of rectangles and using the fact that $f$ is nonnegative and decreasing. The result follows by letting $n\to \infty$ and using the squeeze theorem.

In our problem we have $f(x) = \ln x.$ So $f$ is monotone and the convergence of $\int_a^1 \ln x\, dx$ to $-1$ as $a\to 0^+$ follows from the fact that $a\ln a \to 0.$ Therefore the desired limit follows from the claim.

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There is actually surprisingly one quite simple way to find the limit in this particular case using L'Hospitals.

$$\lim_{n\to\infty} \frac{\sum_{k=1}^n \log(\frac{k}{n})}{n} = -\frac{\infty}{\infty}$$

As you can see this is already in L'hospitals format. Taking the derivative of the top and bottom to find the limit becomes. $$\lim_{n\to\infty}\frac{\sum_{k=1}^n -\frac{1}{n}}{1}$$

And because the Sum goes through the sum 1 to n times (n times) so it will add $$-\frac{1}{n}$$ n times. So the limit becomes $$\lim_{n\to\infty} n(-\frac{1}{n}) = -1$$ From this you can prove that the limit is equal to -1.

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