Mathematics Asked by user834302 on December 16, 2020

The Euler-Mascheroni constant is defined as $gamma := lim_{xtoinfty}(H_n – ln,n)$ where $H_n = sum_{k=1}^{n}frac{1}{k}$, and the de la Vallée-Poussin’s formula states that:

$$gamma = lim_{ntoinfty}frac{1}{n}sum_{k=1}^{n}left(leftlceil frac{n}{k} rightrceil – frac{n}{k}right)$$

How can one prove this formula? If the proof is too long for this site, could you give me the link to an article containing a proof of this fact?

**Update:** I did some work myself. Here it is:

We know that

$$gamma=lim_{ntoinfty}(H_{n}-log n)$$

So we have to prove that

$$lim_{ntoinfty}(H_{n}-log n)=lim_{ntoinfty}frac{1}{n}sum_{k=1}^{n}left(leftlceil frac{n}{k} rightrceil – frac{n}{k}right)$$

Now

begin{align}

lim_{ntoinfty}frac{1}{n}sum_{k=1}^{n}left(leftlceil frac{n}{k} rightrceil – frac{n}{k}right)&=lim_{ntoinfty}frac{1}{n}sum_{k=1}^{n}leftlceil frac{n}{k} rightrceil-lim_{ntoinfty}frac{1}{n}sum_{k=1}^{n}frac{n}{k}\

&=lim_{ntoinfty}frac{1}{n}sum_{k=1}^{n}leftlceil frac{n}{k} rightrceil-lim_{ntoinfty}frac{H_n}{n}

end{align}

So we have to prove that

$$lim_{ntoinfty}(H_{n}-log n)=lim_{ntoinfty}frac{1}{n}sum_{k=1}^{n}left lceil{frac{n}{k}}right rceil-frac{H_n}{n}$$

$$implies lim_{ntoinfty}(nH_{n}-log n^{n})=lim_{ntoinfty}sum_{k=1}^{n}left lceil{frac{n}{k}}right rceil-H_n$$

$$implies lim_{ntoinfty}((n+1)H_{n}-log n^{n})=lim_{ntoinfty}sum_{k=1}^{n}left lceil{frac{n}{k}}right rceil$$

How can I proceed further? Please mention in the comments, did I do something wrong?

Any help would be appreciated.

With your current approach, you would have to split along the integers where $leftlceil frac{n}{k} rightrceil$ is constant, which is unnecessarily complex.

Instead, using Riemann summation, the limit $$lim_{n to infty} frac{1}{n} sum_{k=0}^n left( leftlceil frac{n}{k} rightrceil - frac{n}{k} right)$$

is $$int_0^1 left( leftlceil frac{1}{x} rightrceil - frac{1}{x} right) mathrm{d}x$$

More generally, $lim_{n to infty} frac{1}{n} sum_{k=0}^n fleft( frac{k}{n} right) = int_0^1 f(x) dx$. From here, split it into sections where $leftlceil frac{1}{x} rightrceil = n$. Equivalently, this means that $$n-1 < frac{1}{x} le n to frac{1}{n} le x < frac{1}{n-1}$$ The integral would then be $$sum_{n=2}^{infty} int_{frac{1}{n}}^{frac{1}{n-1}}left( n - frac{1}{x} right) mathrm{d}x$$

Evaluating the integral makes it $$sum_{n=2}^{infty}left(nleft(frac{1}{n-1}-frac{1}{n}right)-lnleft(frac{1}{n-1}right)+lnleft(frac{1}{n}right)right)$$

After simplifying a bit, it is $$sum_{n=2}^{infty}left(frac{1}{n-1}+lnleft(n-1right)-lnleft(nright)right) = sum_{n=1}^{infty}left(frac{1}{n}+lnleft(nright)-lnleft(n+1right)right)$$

Let the upper bound on the sum be $x$. Then this is $$lim_{x to infty} sum_{n=1}^{x}left(frac{1}{n}+lnleft(nright)-lnleft(n+1right)right)$$

The first term is $H_x$ and the rest telescopes to $ln(1) - ln(x+1) = -ln(x+1)$. Altogether, the limit becomes $$lim_{x to infty} left( H_x - ln(x+1) right)$$

You can then add $lim_{x to infty} left( ln(x+1)-ln(x) right) = 0$ to get $$lim_{x to infty} left( H_x - ln(x+1) right) + lim_{x to infty} left( ln(x+1)-ln(x) right) = lim_{x to infty} left( H_x-ln(x) right)$$

which is exactly the original definition of the Euler-Mascheroni constant.

Correct answer by Varun Vejalla on December 16, 2020

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