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