How to prove $x_n$ converges as $n \to \infty$.

Question

Let $x_n = \sum_{k=1}^{n} \frac{1}{k} - \log n$. Prove that $x_n$ converges as $n \to\infty$.

The (finite) limit $\gamma$ of the sequence ${x_n}$ is well-known: it is the Euler constant. Check here: https://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant — Ángel Valencia, Dec 01 '15 at 01:55
@ÁngelValencia So if the limit is well known, why would this be false? — 5xum, Dec 01 '15 at 01:56
Try using left and right Riemann sums to bound $\sum_{k=1}^n \frac{1}{k}$ by something involving $\log (n)$ (since $\int_1^n \frac{1}{x}dx= \log(n)$). — kccu, Dec 01 '15 at 02:01

score 4 · Accepted Answer · answered Dec 01 '15 at 02:20

Hint: A proof without using the integration concepts.

Show that $$\frac{1}{n + 1} < \log\left(1 + \frac{1}{n}\right) < \frac{1}{n}. \tag{1}$$ You might need the facts that the sequence $\left(1 + \frac{1}{n}\right)^n$ increasingly converges to $e$ and the sequence $\left(1 + \frac{1}{n}\right)^{n + 1}$ decreasingly converges to $e$.
Use $(1)$, show that $$\frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n + 1} < \log(n + 1) < 1 + \frac{1}{2} + \cdots + \frac{1}{n}. \tag{2}$$
Use $(2)$, show that the sequence $\{x_n'\}$ is increasing and bounded above, where $$x_n' = 1 + \frac{1}{2} + \cdots + \frac{1}{n} - \log (n + 1).$$ Hence $x_n'$ converges.
Show that $x_n$ and $x_n'$ have the same limit.

Good outline. An alternative route is to show that $x_n' = \int_{1}^{n+1}\left[\frac{1}{\lfloor x\rfloor} - \frac{1}{x}\right]{\rm d}x$ and bound the integrand above to conclude that $x_n'$ converges. — Kibble, Dec 01 '15 at 02:25

1 Answers1