$a_k=\begin{cases}\frac{1}{(k-1)^2} - \frac{1}{k}, n \text{ is odd }\\ \frac{1}{k-1} - \frac{1}{k^2}, k \text{ is even }\end{cases}$
I have to show the series is conditionally convergent, i.e; the series itsrlf is convergent but not absolutely. I'm really unable to show how to prove this. I mean I even tried at least to show that the series isn't absolutely convergent but there are only $2$ test in my book to check absolutely convergent, namely root test, ratio test, but by using these two I fail to conclude anything, even how to show the series is infact convergent?
I'm requesting please I want details solution in this case, please give me answer in answer section rather comment, it would be great help.
Thanks in advance.
If $\sum |a_n| <\infty$ then $\sum |a_{2n}| <\infty$. But $\sum |a_{2n}|=\sum (\frac 1 {2n-1} -\frac 1 {4n^{2}})=\infty$ because $\sum \frac 1 {2n-1}=\infty$ (by comparison with the harmonic series) and $\sum \frac 1 {4n^{2}}<\infty$. Hence the series is not absolutely convergent.
Conditional convergence is very easy and you can do it yourself. Write down the partial sums explicitly. You see lots of cancellations and you will get the parial sums in the form $s_n=1+c_n$ with $c_n \to 0$. (I will let you write down what exactly $c_n$ is). Hence $\sum a_n$ converges with sum $1$.