$$ \mbox{Why is}\quad\lim_{x\to\infty} \frac{\sum_{i = 1}^{x}\left[\sum_{j = 1}^{i}1/j -\ln\left(i\right)-\gamma\right]}{\sum_{i = 1}^{x}1/i} = \frac{1}{2}\ ?.$$

I learnt Euler's Constant $\gamma$ before, and I want to know the sum of $H_k-\ln k-\gamma$. As Wikipedia says, $H_k=\ln k+\gamma+\varepsilon_k$, where $\varepsilon_k\sim\frac1{2k}$. But I wonder how to prove it. By the following Mathematica program, I can check that this limit is $\frac12$. But I cannot prove it. Why is it not anything below or above $\frac12$?

Limit[Sum[Sum[1/j, {j, 1, i}] - Log[i] - EulerGamma, {i, 1, x}]/Sum[1/i, {i, 1, x}], x -> Infinity]
  • 2
    $\begingroup$ I remembered someone has already answered you that you can use Stolz-Cesaro's lemma to answer your question. As far as I observe, his hint is correct, have you tried it? $\endgroup$ Jun 19, 2021 at 23:41
  • $\begingroup$ @ParesseuxNguyen I see. That means the limit I want to calculate equals $\lim\limits_{x\to\infty}\frac{\sum_{i=1}^x\frac1i-\ln x-\gamma}{\frac1x}$, and as wiki says, $H_k=\ln k+\gamma+\varepsilon_k$, where $\varepsilon_k\sim\frac1{2k}$, so that equals $\lim\limits_{x\to\infty}\frac x{2x}=\frac12$. $\endgroup$ Jun 20, 2021 at 5:22
  • $\begingroup$ Also note that reverse Cesaro stolz doesn't always hold true. $\endgroup$ Jun 20, 2021 at 6:57
  • 1
    $\begingroup$ @Alex-Github-Programmer You may find this more general result interesting as well: math.stackexchange.com/q/3551025 $\endgroup$
    – Gary
    Jun 20, 2021 at 10:38

1 Answer 1


With the help of @ParesseuxNguyen, I try to solve it with Stolz-Cesaro's lemma:

$\quad\lim\limits_{x\to\infty}\frac{\sum_{i=1}^x(\sum_{j=1}^i\frac1j-\ln i-\gamma)}{\sum_{i=1}^x\frac1i}\\ =\liminf\limits_{i\to\infty}\frac{\sum_{j=1}^i\frac1j-\ln i-\gamma}{\frac1i}\\ =\limsup\limits_{i\to\infty}\frac{\sum_{j=1}^i\frac1j-\ln i-\gamma}{\frac1i}\\ =\lim \limits_{i\to\infty}\frac{\sum_{j=1}^i\frac1j-\ln i-\gamma}{\frac1i}\\ =\lim \limits_{i\to\infty}\frac i{2i}=\frac12$.

(Since $H_i-\ln i-\gamma\sim\frac1{2i}$, so the last equality is right.)

  • $\begingroup$ Why is $H_i-\log(i)-\gamma\sim\frac1{2i}$? $\endgroup$
    – robjohn
    Jul 4, 2021 at 17:38

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