calculus - Given the parameter $a in mathbb{R}$ such that the given sequence $(x_n)_{n ge 1}$ is bounded, find the limit of the sequence $x_n$.

I am given the following sequence $(x_n)_{n \ge 1}$:

$x_n = \bigg ( 1 + \dfrac{1}{3}
+ \dfrac{1}{5} + ... + \dfrac{1}{2n-1}\bigg ) - a \bigg ( 1 + \dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{n} \bigg )$

With $a \in \mathbb{R}$ such that the sequence $x_n$ is bounded. I am asked to find the limit of the sequence $x_n$.

I tried completing the sum:

$$x_n = \bigg ( 1 + \dfrac{1}{2}
+ \dfrac{1}{3} + \dfrac{1}{4} + ... + \dfrac{1}{2n-1} + \dfrac{1}{2n} \bigg ) -
\bigg ( \dfrac{1}{2} + \dfrac{1}{4} + ... + \dfrac{1}{2n} \bigg ) - a \bigg ( 1 + \dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{n} \bigg )$$

$$x_n = \bigg ( 1 + \dfrac{1}{2}
+ \dfrac{1}{3} + \dfrac{1}{4} + ... + \dfrac{1}{2n-1} + \dfrac{1}{2n} \bigg ) - \dfrac{1}{2}
\bigg ( 1 + \dfrac{1}{2} + ... + \dfrac{1}{n} \bigg ) - a \bigg ( 1 + \dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{n} \bigg )$$

$$x_n = \bigg ( 1 + \dfrac{1}{2}
+ \dfrac{1}{3} + \dfrac{1}{4} + ... + \dfrac{1}{2n-1} + \dfrac{1}{2n} \bigg ) -
\bigg ( a + \dfrac{1}{2} \bigg ) \bigg ( 1 + \dfrac{1}{2} + ... + \dfrac{1}{n} \bigg ) $$

And here I tried using the fact that:

$$\sum\limits_{k=1}^\infty \dfrac{1}{k} \approx \ln n$$

(I have only used this thing once before, so I'm expecting to use it incorrectly. I know it is called the Harmonic number). Using this I got into a whole mess with logarithms and what not. I arrived at a pretty random answer, while my textbook claims the right answer is $\ln 2$. How should I solve this exercise, arriving at $\ln 2$ ?

Answer

\begin{align*}
x_{n}=\left(1-\left(a+\dfrac{1}{2}\right)\right)\left(\sum_{k=1}^{n}\dfrac{1}{k}\right)+\left(\dfrac{1}{n+1}+\cdots+\dfrac{1}{2n}\right),
\end{align*}
and we have
\begin{align*}
\dfrac{1}{n+1}+\cdots+\dfrac{1}{2n}\rightarrow\int_{0}^{1}\dfrac{1}{1+x}dx=\log 2,
\end{align*}
so this forces $1-(a+1/2)=0$ and the limit is $\log 2$.