Showing Convergence of Sequence from monotone subsequences My questions is as follows:
Let $(u_n)$ be defined by $u_1=1$ and
$$u_{n+1} = u_n + \frac{(-1)^n}{n} \;\;\;\; \forall n \in \mathbb{N}$$
Show that the subsequence $(u_{2k-1})_{k\in\mathbb{N}}$ is monotone decreasing and that the subsequence $(u_{2k})_{k\in\mathbb{N}}$ is monotonic increasing. Conclude from this that the sequence $(u_{n})_{n\in\mathbb{N}}$ is convergent.

So far I have:
$$\begin{aligned}
u_{2k-1} &= u_{2k-2} + \frac{(-1)^{2k-2}}{2k-2} \\
&=u_{2k-2} + \frac{1}{2k-2} \\
&=u_{2k-3}  + \frac{(-1)^{2k-3}}{2k-3}+\frac{1}{2k-2} \\
&=u_{2(k-1)-1} -\frac{1}{(2k-3)(2k-2)} \\
u_{2k-1} - u_{2(k-1)-1} &= -\frac{1}{(2k-3)(2k-2)} <0
\end{aligned}$$
Hence $u_{2k-1} < u_{2(k-1)-1}$ and so $(u_{2k-1})$ is decreasing. I used a similar argument to show that $(u_{2k})$ is increasing. I have thought for a while on how to conclude convergence, and one way to go about it would be to show $(u_{n})$ is Cauchy, but I am not sure how to go about doing this.
Any help would be much appreciated. Also please bear in mind I am quite new to analysis, so may have overlooked something obvious.
 A: There are technically two ways to go about proving that the sequence $\{u_n\}_{n=1}^\infty$ converges. Since you say that you are new to analysis, I will note that either argument is perfectly valid as far as you should be concerned. Though it might be to your benefit to notice that the classical argument explicitly invokes a limit which in a more general context we won't be able to know. I say more general context as clearly,
\begin{aligned} \lim_{n \to \infty} u_n &=1 - \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}  \\&=1-\log(2). \end{aligned} 


*

*Classical: Show that the subsequence $\{u_{2n-1}\}_{n=1}^\infty$ is bounded below and decreasing. Show that the subsequence $\{u_{2n}\}_{n=1}^\infty$ is bounded above and increasing. Conclude both subsequences are convergent sequences by the monotone convergence theorem for sequences which implies the nonconstructive proposition known as the limited principle of omniscience. Conclude both subsequences converge to a common limit.

*Constructive: Without loss of generality, suppose $n>m$. You may prove (an exercise left for the reader) via strong induction $(*)$ that for every $n \in \mathbb N$, we have
$$0 \leq \frac1{m+1}-\frac1{m+2}+\frac1{m+3}-\cdots \pm \frac1{n}\leq \frac1{m+1}\,.
$$
Let $\varepsilon>0$ be given. By the Archimedean Principle, we may find a positive integer $N$ so that $N>\frac1{\varepsilon}$. So for all $n>m\geq N$, we have
\begin{aligned} \left|u_n - u_m \right| &= \left|\frac1{m+1}-\frac1{m+2}+\frac1{m+3}-\cdots \pm \frac1{n}\right| \\& \leq \frac1{m+1} \\& < \varepsilon. \end{aligned}Therefore, the sequence $\{u_n\}_{n=1}^\infty$ is a Cauchy sequence.


$(*)$ Hint: There are two cases to consider for the induction step, even and odd.
A: For any $k$  we have $u_1\geq u_{2k-1}>u_{2k}$ so the increasing sequence $(u_{2k})_k$ is bounded above by $u_1$ so it has a limit $A.$
For any $k$  we have  $u_2\leq u_{2k}<u_{2k+1}$ so the decreasing sequence $(u_{2k+1})_k$ is bounded below by $u_2$ so it has a limit $B.$
We have $A-B=(\lim_{k\to \infty}u_{2k})-(\lim_{k\to \infty}u_{2k+1})=\lim_{k\to \infty}(u_{2k}-u_{2k+1})=0$. Therefore $A=B.$
Given $e>0,$ take $k_1\in \Bbb N$ such that $k\geq k_1\implies |u_{2k}-A|<e$ and take $k_2\in\Bbb N$ such that $k\geq k_2\implies |u_{2k+1}-A|<e.$ Then $n\geq 1+2\max (k_1,k_2)\implies |u_n-A|<e.$
