Find the limit $\lim\limits_{n \to \infty} \left ( \left (1 + \frac {1} {2} + \cdots + \frac {1} {n} \right ) - \ln n \right ).$

Question

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.

Answer 1

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 $n$.

Answer 2

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...