<s>[INST] Limit of $ (x_{n}) = ( 1 + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{2n-1} ) - a ( 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} ) $ Let $ a \in \mathbb{R} $ so that the sequence $ (x_{n})_{n\geq 1} $
$ (x_{n}) = ( 1 + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{2n-1} ) - a ( 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} ) $ is bounded.
Find $\lim_{n\to\infty}  (x_{n})  $
If $ (x_{n}) $ is bounded that means there are $ 2 $ real numbers $m$ and $q$ let's say so that $ m \leq (x_{n}) \leq q $
Is there a more simple way to approach this problem? , other than writing things like :
$  \frac{\sum_{k=1}^{n}\frac{1}{2k-1}-m}{\sum_{k=1}^{n}\frac{1}{k}} \geq a \geq \frac{\sum_{k=1}^{n}\frac{1}{2k-1}-q}{\sum_{k=1}^{n}\frac{1}{k}}$ 
The correct answer should be $ ln2 $
 [/INST] Let me give you a sketch of the argument.
When you increase $n$ by $1$, what happens?  In net,
$$
x_{n+1}-x_n=\frac{1}{2(n+1)-1}-a\left(\frac{1}{n+1}\right)=\frac{1}{2n+1}-\frac{a}{n+1}.
$$
Using this, you can show that it must be the case that $a=\frac{1}{2}$.  Why?  If $a=\frac{1}{2}+\delta$ where $\delta>0$, then
$$
x_{n+1}-x_n=\frac{1}{2n+1}-\frac{a}{n+1}=\frac{1}{2n+1}-\frac{1}{2n+2}-\frac{\delta}{n+1}=-\frac{\delta}{n}+O\left(\frac{1}{n^2}\right),
$$
and so the tail of the sum behaves like $-\delta\sum\frac{1}{n}$ (which must diverge to $-\infty$).  Similarly, you can show that if $a=\frac{1}{2}-\delta$ where $\delta>0$, then the series must diverge to $+\infty$.
Now, we know that $a=\frac{1}{2}$. Now, note that we can write
$$
\begin{align*}
x_n&=\left(1+\frac{1}{3}+\cdots+\frac{1}{2n-1}\right)-\frac{1}{2}\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)\\
&=\left(1+\frac{1}{2}+\cdots+\frac{1}{2n}\right)-\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2n}\right)-\frac{1}{2}\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)\\
&=\left(1+\frac{1}{2}+\cdots+\frac{1}{2n}\right)-\frac{1}{2}\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)-\frac{1}{2}\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)\\
&=\left(1+\frac{1}{2}+\cdots+\frac{1}{2n}\right)-\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)\\
&=y_{2n}-y_n,\qquad y_n:=1+\frac{1}{2}+\cdots+\frac{1}{n}.
\end{align*}
$$
Now, we know (or if not, you can show) that as $n\to\infty$, $y_n=\ln n+\gamma+O\left(\frac{1}{n}\right)$, where $\gamma$ is the Euler–Mascheroni constant, so that
$$
x_n=y_{2n}-y_n=(\ln(2n)+\gamma)-(\ln(n)+\gamma)+O\left(\frac{1}{n}\right)\to\ln2\text{ as }n\to\infty.
$$
</s>