Test for the convergence of the series $\sum _{_{n=1}}^{\infty }\frac{\frac{1}{2}+\left(-1\right)^n}{n}$ Test for the convergence of the series $\sum _{_{n=1}}^{\infty }\frac{\frac{1}{2}+\left(-1\right)^n}{n}$
My attempt
$\sum _{_{n=1}}^{\infty }\frac{\frac{1}{2}+\left(-1\right)^n}{n}=\frac{-\frac{1}{2}}{1}+\frac{\frac{3}{2}}{2}+\frac{-\frac{1}{2}}{3}+\frac{\frac{3}{2}}{4}+\frac{-\frac{1}{2}}{5}+\frac{\frac{3}{2}}{6}+...$. I tried to use Drichlet test, ratio test.
I applied limit comparison test with $a_n=\frac{1}{n}$. $\lim_{n\to \infty}\frac{\frac{\frac{1}{2}+\left(-1\right)^n}{n}}{\frac{1}{n}}=\lim_{ n \to \infty}\frac{1}{2}+\left(-1\right)^n\neq 0(\because$ sequence doesnot converges to anyone of the real number, I can prove it). So, series diverges. 
 A: Note that your series is the sum of a divergent series and a convergent series, so it diverges. More specifically: ( I will write $\sum$ for $\sum_{n=1}^{\infty}$ ) $$\sum\frac{\frac{1}{2}+(-1)^n}{n}=\sum\frac{\frac{1}{2}}{n}+\sum\frac{(-1)^n}{n}=\frac{1}{2}\sum\frac{1}{n}+\sum\frac{(-1)^n}{n}$$
Observe the first sum there is half the harmonic sum, which is known to diverge. The second is the alternating harmonic sum, which is convergent by the Alternating Series Test. 
A basic result is that $$\text{If}\space \sum a_n\space\text{converges and} \sum b_n\space\text{diverges, then}\space\sum a_n+b_n\space\text{diverges.}\space$$ Apply this result. 
A: No! $\lim \frac{a_n}{b_n}$ does not exist doesn't allow you to conclude anything.
As an example, let $b_n=\frac1n$ the harmonic series, and let
$$
\frac{a_n}{b_n}=\begin{cases}
\sqrt{n} & n\text{ is of the form }2^k, k\in\mathbb{N}\\
\frac1{\sqrt{n}} & \text{otherwise}
\end{cases}
$$
then $\lim\frac{a_n}{b_n}$ does not exist, but $\sum a_n$ converges.
A: Concatenate terms in pairs, with the odd n first.  $(\frac{-1}{2n}+\frac{3}{2(n+1)})=\frac{2n-1}{2n^2}\gt \frac{1}{2n}$  The sum then becomes $\gt \sum_{k=1}^\infty \frac{1}{2(1+2k)}$ which diverges. 
