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 $
 A: Let 
$$x_n(a)=\left(1+{1\over3}+{1\over5}+\cdots+{1\over2n-1}\right)-a\left(1+{1\over2}+{1\over3}+\cdots+{1\over n}\right)$$
If $|x_n(a)|$ and $|x_n(b)|$ are both bounded as $n\to\infty$, then so is 
$$|x_n(a)-x_n(b)|=|a-b|\left(1+{1\over2}+{1\over3}+\cdots+{1\over n}\right)$$ 
which is not true if $a\not=b$. Hence there can be at most one $a\in\mathbb{R}$ for which $|x_n(a)|$ is bounded as $n\to\infty$.  
It turns out that $|x_n(1/2)|$ is bounded: As in Nick Peterson's answer,
$$\left(1+{1\over3}+{1\over5}+\cdots+{1\over2n-1}\right)-{1\over2}\left(1+{1\over2}+{1\over3}+\cdots+{1\over n}\right)\\
=\left(1+{1\over2}+{1\over3}+\cdots+{1\over2n}\right)-\left(1+{1\over2}+{1\over3}+\cdots+{1\over n}\right)\\
={1\over n+1}+{1\over n+2}+\cdots+{1\over n+n}\\
={1\over n}\sum_{k=1}^n{1\over1+k/n}\to\int_0^1{dx\over1+x}=\ln2$$
The final step(s) require knowing the natural logarithm as an integral, and that an integral is a limit of Riemann sums.
A: 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
$$
x_n=y_{2n}-y_n=(\ln(2n)+\gamma)-(\ln(n)+\gamma)+O\left(\frac{1}{n}\right)\to\ln2\text{ as }n\to\infty.
$$
A: Let $H_n=1+\dots+\frac{1}{n}.$ For $x_n$ to be bounded, $a$ has to be positive.
It is $$x_n=H_{2n}-aH_n-(\frac{1}{2}+\dots+\frac{1}{2n})=H_{2n}-(a+\frac{1}{2})H_n.$$ Assume $a\neq1/2$, let's say $a=\frac{1}{2}\pm\varepsilon$, where $\varepsilon$ is a positive quantity. Then $x_n=H_{2n}-(1\pm\varepsilon) H_n$. 
Later on, we will Prove that $H_{2n}-H_n$ converges to $\log(2)$. If we use it now though, we get that $H_{2n}-H_n$ is bounded and $\pm\varepsilon H_n=(H_{2n}-H_n)-x_n$, therefore $(H_n)$ is bounded (note that $\varepsilon$ is fixed!) which is a contradiction, since $H_n$ is well-known to be divergent. So $a=\frac{1}{2}$ and all we have to do is prove that $H_{2n}-H_n\to\log(2)$ as promised.
First observe (maybe use induction if you want a detailed proof) that $H_{2n}-H_n=\displaystyle{\sum_{k=1}^{2n}\frac{(-1)^{k-1}}{k}}:=s_{2n}$, with $s_n$ being the partial sums of the series $\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}$, which is a convergent series by the Leibniz alternating-terms criterion. Therefore, the subsequence $(s_{2n})$ of $(s_n)$ converges as well and to the same limit too: this limit is exactly $\log(2)$, by the Taylor expansion of $x\mapsto\log(1+x)$ for $|x|\leq1$.
