<s>[INST] Discuss the convergence of the series $\sum_{n=1}^{\infty}\frac{1/2+(-1)^n}{n}$ 
Discuss the convergence of the  series $\displaystyle\sum_{n=1}^{\infty}\frac{\frac{1}{2}+(-1)^n}{n}$.

Attempt: Since the given series is alternating series, I am using Leibniz test.
The terms of the series are $\frac{-1}{2\cdot 1},\frac{3}{2\cdot 2},\frac{-1}{2\cdot 3},\frac{3}{2\cdot 4},\frac{-1}{2\cdot 5},\ldots$
That is equal to $\frac{1}{2}(-1,\frac{3}{2},\frac{-1}{3},\frac{3}{4},\frac{-1}{5},\ldots)$.
Hence $a_n= \frac{-1}{2n-1}$ if $n$ is odd and $a_n=\frac{3}{n}$ if $n$ is even.
$a_n$ goes to zero. But $a_n$ not decreasing. Hence the series divergent.This argument will work or not please help me!
 [/INST] I would look at creating a sequence that consists of summing 2 subsequent terms of the original sequence.
$b_{n} = a_{2n+1} + a_{2n+2} = \frac{-\frac{1}{2}}{2n + 1} + \frac{\frac{1}{2}}{2n+ 2}
 = \frac{1}{2}(\frac{1}{2n+2} - \frac{1}{2n + 1}) = \frac{1}{2}\frac{(2n+1)-(2n+2)}{(2n+1)(2n+2)} = -\frac{1}{2}\frac{1}{(2n+1)(2n+2)}$
I'm not sure about the actual value but if I am correct the series of $b_n$ converges and therefor the series of $a_n$ converges.
Edit: Sorry, I made an error. My the series of my sequence $b_n$ as constructed above is convergent but is not the same as yours.
The correct one is:
$b_n = a_{2n+1} + a_{2n+2} = \frac{\frac{1}{2}}{2n+1} + \frac{\frac{3}{2}}{2n+2} = \frac{1}{2}\frac{4n+1}{(2n+1)(2n+2)}$
Which is divergent.
</s>