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Prove that the following limit exists and find the limit $:$ $$\lim\limits_{n \to \infty} \left ( \left (1 + \dfrac {1} {2} + \cdots + \dfrac {1} {n} \right ) - \ln n \right ).$$

I know that the above sequence is strictly decreasing and bounded below by $0$ and hence by monotone convergence theorem it has to converge to it's infimum which is known as Euler-Mascheroni's constant. But how do I evaluate the limit?

Any help or suggestion in this regard will be highly appreciated. Thanks in advance.

Source $:$ ISI (Indian Statistical Institute) PhD entrance test in Mathematics, TEST CODE : MTA (FORENOON SESSION) (Question No. $1$) held in $20$th September this year.

Anacardium
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  • What do you mean by evaluate? Up to what accuracy? – pancini Nov 01 '20 at 06:10
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    The question is badly phrased. You cannot 'find' the limit. You can find approximate values. The limit is the Euler constant $\gamma$. – Kavi Rama Murthy Nov 01 '20 at 06:11
  • That (The degree of accuracy) is not specified in the question. How do I know @Elliot G? – Anacardium Nov 01 '20 at 06:11
  • @user2661923 OP stated in the question that they are already aware of the constant – angryavian Nov 01 '20 at 06:17
  • @angryavian You're right. I glossed over that, when I saw "But how do I evaluate the limit". – user2661923 Nov 01 '20 at 06:20
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    The limit is given a name and a symbols because it cannot be evaluated in terms of other known constants. So the question is wrong. I am not saying that you copied it wrongly but I am saying that the question given in the Entrance Exam is itself wrong. – Kavi Rama Murthy Nov 01 '20 at 07:39
  • @KaviRamaMurthy sir I apologize for the misinterpretation of your statement. Sir I request you to kindly allow me a space to ask you another question which is - "Will the examinees be awarded full marks if he/she attempted the question and did the first part (i.e. showing the convergence of the given sequence) correctly?" Hoping to get a reply from your end regarding this as soon as possible. Thank you very much sir. – Anacardium Nov 01 '20 at 08:07
  • This topic has already been covered in multiple threads on this website. Please do your research before asking. – K.defaoite Nov 01 '20 at 10:46
  • @K.defaoite the link you have provided is very helpful. In fact robjohn's answer is something that I was looking for. Thanks for the link. BTW why is the downvote? I have seen persons in this website for the last couple of weeks who are deliberately downvoting my post without any reason. Is there any way to find those unscrupulous persons? Thanks again. – Anacardium Nov 01 '20 at 15:00

2 Answers2

1

From the definition of the Riemann's integral, it is easy to see that: $$ \int_1^{n+1}\frac1xdx\lt\sum_{k=1}^n\frac 1k\lt1+\int_1^{n}\frac1xdx $$ We arrive at $$ \ln(n+1)\lt\sum_{k=1}^n\frac 1k\lt1+\ln n$$ As a rough approximation we can take the average of the 2 bounds: $$ \sum_{k=1}^n\frac 1k\approx\frac12+\frac12\ln [n(n+1)] $$ and the appoximation for $\gamma$ is: $$ \gamma\approx\frac12+\frac12\ln [n(n+1)]-\ln n=\frac12+\frac12\ln(1+\frac1n) $$ As $n$ increases we get $\gamma\approx0.5$. The actual value is $\gamma=0.577...$ so it is not highly accurate, but considering the effort (not much) I think it should be satisfying...

am301
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0

It is known that $$ 1 + \frac{1}{2} + \ldots + \frac{1}{n} - \log n = \gamma + \frac{1}{{2n}} - \sum\limits_{k = 1}^{N - 1} {\frac{{B_{2k} }}{{2k}}\frac{1}{{n^{2k} }}} - \theta _N (n)\frac{{B_{2N} }}{{2N}}\frac{1}{{n^{2N} }} $$ where $N\ge1$, $0<\theta_N(n)<1$ is a suitable number and the $B_m$ denote the Bernoulli numbers. You can re-arrange this formula to obtain accurate approximations for $\gamma$. If you take $N = \left\lfloor {\pi n} \right\rfloor$, $$ \left| {\frac{{B_{2N} }}{{2N}}\frac{1}{{n^{2N} }}} \right| = \mathcal{O}\!\left(\frac{\mathrm{e}^{ - 2\pi n}}{\sqrt n} \right). $$ Hence, the error will be exponentially small in $n$.

Gary
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