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Suppose $\gamma$ is the Euler–Mascheroni constant .

Let $\gamma_n:=H_n-\ln(n)$.

Then $\gamma=\lim_{n \to \infty} \gamma_n.$

I have this summation $$\sum_{n=1}^{\infty}(\gamma_n-\gamma)$$

I would like to know if this summation converges or not.

If I calculate the limit $\lim_{n \to \infty}(\gamma_n-\gamma)$ it is clear that the result is zero. but it seems to me it diverges, so how to find the result if it converges and how to prove it.

EDIT

Now we are sure it diverges by using the harmonic formula shown in Iqcd answer.

But I think if we speed up $\gamma_n$ to go quickly to $\gamma$ it will converges by changing $n$ to $n^2$ so we have $\sum_{n=1}^{\infty}(\gamma_{(n^2)}-\gamma)$.

And it seems we have new function here diverges at $x=1$ $$f(x)=\sum_{n=1}^{\infty}(\gamma_{(n^x)}-\gamma)$$

at $x \to +\infty$ , $f(x)=1-\gamma$

Did
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Pentapolis
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    To make your question better, what is $H_n$?. "but it seems to me it diverges", why do you think so? –  Oct 09 '16 at 12:21
  • Me too, but my question was to know how to prove it. – Pentapolis Oct 09 '16 at 12:37
  • May be it is interesting, that $\displaystyle \lim_{n \to \infty} \sum_{k=4^n}^{4^{n+1}-1} (H_k - \log(k) -\gamma) \to \log(2) $ (heuristically...) – Gottfried Helms Oct 12 '16 at 23:49
  • @GottfriedHelms, it seems we are close to something see this http://math.stackexchange.com/questions/1912650/an-easy-calculated-limit-looks-like-the-one-of-euler-mascheroni-constant – Pentapolis Oct 12 '16 at 23:56

1 Answers1

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We have (see Wikipedia)

$$\tag{$*$}\label{eq1}H_n=\log(n)+\gamma+\frac1{2n}+O(n^{-2})$$

so

$$\sum_{n\leq x}\left\{H_n-\log(n)-\gamma\right\} =\sum_{n\leq x}\left\{\frac1{2n}+O(n^{-2})\right\},$$

and therefore the series diverges.

ADDED (after the edit): let us define the function

$$f(x;N):=\sum_{n\leq N}\left\{H_{n^x}-\log(n^x)-\gamma\right\},$$

so that $f(x;N)\to f(x)$ when $N\to\infty$. Again by \eqref{eq1} we see that

$$f(x;N)=\sum_{n\leq N}\left\{\log(n^x)+\gamma+\frac1{2n^x}-\log(n^x)-\gamma+O(n^{-2x})\right\} =\sum_{n\leq N}\left\{\frac1{2n^x}+O(n^{-2x})\right\}.$$

Since $\sum n^{-p}$ converges iff $p>1$, we see that $f(x)$ converges iff $x>1$.

user246336
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  • If you don't mind see the edit. – Pentapolis Oct 09 '16 at 14:04
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    I've added the convergence of $f(x)$. Is that what you asked? – user246336 Oct 09 '16 at 14:18
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    $H_n = \int_1^{n+1} \frac{1}{\lfloor x \rfloor}dx$ so $\log(n)-H_n= \int_1^{n+1} (\frac{1}{x}-\frac{1}{\lfloor x \rfloor})dx$ and $-\gamma = \int_1^\infty (\frac{1}{x}-\frac{1}{\lfloor x \rfloor})dx$ while $\gamma_n-\gamma =\int_{n+1}^\infty (\frac{1}{x}-\frac{1}{\lfloor x \rfloor})dx =\int_{n+1}^\infty \int_{\lfloor x \rfloor}^{x} \frac{-2}{y^2} dydx = \int_{n+1}^\infty (\frac{-1}{y^2} + \mathcal{O}(\frac{1}{y^3}))dy = \frac{1}{2n+1}+\mathcal{O}(\frac{1}{n^3})$ @Pentapolis – reuns Oct 09 '16 at 14:20
  • @iqcd, yes that is what I was asking about. so the function $f(x)$ has a real curve for $1<x<+\infty$. – Pentapolis Oct 09 '16 at 16:46