Find : $\lim\_{n\to \infty} (1-\frac{1}{2}+\frac{1}{3}-...-\frac{1}{2n})$ .

**My Approach:**

Let $x\_n=(1-\frac{1}{2}+\frac{1}{3}-...-\frac{1}{2n})$ .

I know that $\gamma\_{n}=\sum\_{n=1}^n \frac{1}{n} -\log(n)$ . And $<\gamma\_n>$ converges to Euler's constant i.e. $\gamma$ which lies between $0.3$ and $1$ . 

Using this fact I found ,

$x\_n=1+\gamma\_{2n-1}+\log(2n-1)-\gamma\_{2n}-\log(2n)$

since $\gamma\_{2n-1}$ and $\gamma\_{2n}$ are subsequences of convergent sequence $<\gamma\_n>$ , so they converges to same limit as $n$ goes to $\infty$ .

Hence , $\lim\_{n\to \infty} x\_n= 1$ .

**Question:**

To check whether I'm wrong or right . Actually I don't have answers , Please help!

**EDIT:**
$(1)$ My approach is wrong.