I'm an undergraduate student in geology and I'm dealing with a project in math. The last question of the project gives me the harmonic series ($A_n = 1 + \frac{1}{2} + ... + \frac{1}{n}$) and this natural logarithm $L = \ln(1+n)$ and asks me to see what happens with the difference $A_n-L$ when $n \to \infty$. I've already done some symbolic computations in Matlab and the answer I get is "eulergamma". So, after some online research I found out that eulergamma is the Euler-Mascheroni constant (being 0.5772....). Not knowing too much about this, I answered the question as follows: "I notice that $\lim\limits_{n \to \infty} (A_n-L)$ behave same as $\lim\limits_{n \to \infty} (A_n-\ln(n))$ which is equal to Euler-Mascheroni constant. So, the requested difference $A_n-L$ is equal to Euler-Mascheroni constant as $n \to \infty$."

Teacher asked me "Why does this $A_n-L$ behave in the same way as this $A_n-\ln(n)$? Try to do something. Add/subtract $\ln(n)$ for example."

Afterall here I am and asking for your help. I tried to add/subtract $\ln(n)$ but nothing came up. Could you please help me with this, or at least give me some clues/hints? I don't have such an experience on this!!

Thank you for your attention!


We have $$A_n-L=\sum_{k=1}^n\frac 1 k-\log(n+1)=\sum_{k=1}^n\frac 1 k\underbrace{-\log(n+1)+\log n}_{=-\log\left(1+\frac 1 n\right)}-\log n\\=\sum_{k=1}^n\frac 1 k-\log n-\log\left(1+\frac 1 n\right)$$

Now since $\displaystyle\lim_{n\to\infty}\log\left(1+\frac 1 n\right)=0$ then

$$\lim_{n\to\infty}\sum_{k=1}^n\frac 1 k-\log n=\lim_{n\to\infty}A_n-L=\gamma$$

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    $\begingroup$ I am grateful for your answer Sami Ben Romdhane! I'm hitting my head right now! The mistake I made was stuck in this form: ln(n)-ln(1+n) = ln(n/(1+n)) instead of doing this: ln(n)-ln(1+n) = -(ln(1+n)-ln(n)) = -ln((1+n)/n) = -ln(1/n+1). Thanks again!! p.s. I can't vote up because I don't have enough reputation points. $\endgroup$ – KonstaNtie Jan 16 '14 at 22:55
  • $\begingroup$ You're welcome. $\endgroup$ – user63181 Jan 16 '14 at 23:01
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    $\begingroup$ @KonstaNtie if you were me you'd be hitting your head because 25 is not divisible by 3. I think head hitting in your case is not appropriate. I'll upvote Sami's answer for you (and me). $\endgroup$ – Betty Mock Jan 16 '14 at 23:51
  • $\begingroup$ @SamiBenRomdhane: Happy Wahdat Days to you dear brother. :+) $\endgroup$ – mrs Jan 17 '14 at 6:35
  • $\begingroup$ Hahaha :P Thank you all!!! $\endgroup$ – KonstaNtie Jan 17 '14 at 19:11

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