Prove that the succession $a_n=\frac{1}{n+1}+\frac{1}{n+2}+\dots \frac{1}{2n}$ is Cauchy or not.

Question

I want to see if this sequence $a_n=\dfrac{1}{n+1}+\dfrac{1}{n+2}+\dots \dfrac{1}{2n}$ is Cauchy.

What I have done:

I want to prove $|a_{n+m}-a_n|<\epsilon $. Using the triangle inequality,

$|a_{n+m}-a_n| =\bigg|\dfrac{1}{2(n+1)}+\dfrac{1}{2(n+2)}+ \dots + \dfrac{1}{2(n+m)}\bigg|\leq \bigg|\dfrac{1}{2(n+1)}\bigg|+\bigg|\dfrac{1}{2(n+2)}\bigg|+ \dots + \bigg|\dfrac{1}{2(n+m)}\bigg| = \dfrac{1}{2}·\bigg(\dfrac{1}{n+1}+\dfrac{1}{n+2} + \dots + \dfrac{1}{n+m}\bigg)$.

And here I dont't know how to continue.

Answer 1

You aren't going to be able to solve it like that, because the series $\frac{1}{n}$ diverges, you need to take into account the sign of the summands.

First notice that $a_{n+1}-a_n=\frac{1}{2n+2}+\frac{1}{2n+1}-\frac{1}{n+1}=\frac{1}{2n+1}-\frac{1}{2n+2}=\frac{1}{(2n+1)(2n+2)}$.

It follows that if $m>N$ then $a_m-a_N\leq \sum\limits_{n=N}^\infty \frac{1}{(2n+1)(2n+2)}$.

Notice that the series $\sum\limits_{n=1}^\infty \frac{1}{(2n+1)(2n+2)}$ converges as it is dominated by $\sum\limits_{n=1}^\infty \frac{1}{n^2}$ which is well-known to converge.

We conclude that the sums of the tails converge to zero, so we are done.

Answer 2

$$\lim_{\infty}a_n=\lim_{\infty}\sum_{k=1}^n\dfrac{1}{1+\frac{k}{n}}\dfrac{1}{n}=\int_0^1\frac{1}{1+x}dx=\ln2$$

Answer 3

$$a_{n+1}-a_n=\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1}$$ $$=\frac{1}{2n+1}-\frac{1}{2n+2}=\frac{1}{(2n+1)(2n+2)}$$

Thus we see that the series is increasing. It converges by a comparison with $\sum \frac{1}{n^2}$. Or by noting that it is bounded above

$$a_n\leq \frac{n}{n+1}<1.$$

We may also use that

$$H_n-\ln n \to \gamma$$ to find that

$$H_{2n}-H_n-\ln(2n)+\ln n\to 0$$ and thus

$$a_n=H_{2n}-H_n\to \ln 2$$