# Finding $\lim_{n \to \infty} 1+ 1/2 + 1/3 +\dots +1/n - \log n$. [duplicate]

$$x_n = 1 + 1/2 +\dots +1/n- \log n$$

Then -

$$1.$$ Is the sequence increasing?

$$2.$$ is the sequence convergent?

For $$(1)$$, $$\sum 1/n$$ is increasing and $$\log n$$ is also increasing. First few terms are increasing, but i don't know about later terms.

$$(2)$$ $$n^{th}$$ term of the sequence can be written as $$a_n = (\sum_{i=1}^{n}) - \log n$$

So, $$\lim_{n\to \infty} a_n = \lim_{n \to \infty} \sum 1/n -\lim_{n \to \infty} \log n$$

Neither first part nor second is convergent here. so i could not conclude anything.

How to solve?

• The sequence is found here: en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant Dec 24, 2018 at 14:49
• Hint: consider that $\log n = \int_{1}^{n} \frac{dx}{x}$. You can obtain the relevant estimates via Riemann sums. Dec 24, 2018 at 14:56
• You have to be careful: $\lim_n (a_n- b_n) = \lim_n a_n - \lim_n b_n$ is not true when $\lim_n a_n =\lim_n b_n=\infty$. Dec 24, 2018 at 16:19
• @Mircea , that's why I left the problem there. Dec 24, 2018 at 16:29

This sequences converges to the Euler–Mascheroni constant. It’s very important in number theory.

Let $$x_n=-\log(n)+\sum_{k=1}^n \frac1k$$. Then, using $$\log(1+x)\ge \frac{x}{1+x}$$, we see that

\begin{align} x_{n+1}-x_n&=\frac1{n+1}-\log\left(1+\frac1n\right)\\\\ &\le \frac1{n+1}-\frac{1}{n+1}\\\\ &=0 \end{align}

and $$x_n$$ is decreasing.

Next, we can estimate the harmonic sum as $$\sum_{k=1}^n \frac1k\ge \frac12 \sum_{k=1}^{n-1}\left(\frac1k+\frac1{k+1}\right)$$, which represents the Trapezoidal Rule approximation of $$\int_1^n \frac1x\,dx$$.

Inasmuch as $$\frac1x$$ is convex, the trapezoidal rule approximation overestimates the integral of $$\frac1x$$ and we have

$$\sum_{k=1}^n\frac1k-\log(n)\ge \sum_{k=1}^n \frac1k -\log(n)-\frac12-\frac1{2n}\ge0$$

whence we see that

$$x_n\ge \frac12$$

Since $$x_n$$ is decreasing and bounded below by $$\frac12$$, the sequence $$x_n$$ converges.