Given the series, $$\sum_{n=1}^{\infty} (-1)^{n}\frac{n}{n+1}$$
I know we can immediately conclude that it is obviously divergent by the divergence test. But I want to know where exactly am I going wrong in the following 'proof' as I have just started learning about convergence and divergence.
Here is what I did :
\begin{align}\sum_{n=1}^\infty (-1)^n \frac{n}{n+1} & = \left(-\frac{1}{2} + \frac{2}{3}\right)+\left(-\frac{3}{4} + \frac{4}{5}\right)+\left(-\frac{5}{6} + \frac{6}{7}\right)+\dots \\ & = \frac{1}{6}+\frac{1}{20}+\frac{1}{42} + \dots\\ & = \frac{1}{2\cdot 3}+\frac{1}{4\cdot 5}+\frac{1}{6\cdot 7}+\dots \\ & = \sum_{n=1}^\infty\frac{1}{(n+1)(n+2)}-\left(\frac{1}{3\cdot 4}+\frac{1}{5\cdot 6}+\frac{1}{7\cdot 8}+\dots\right) \\ & = \sum_{n=1}^\infty\frac{1}{(n+1)(n+2)}- \sum_{n=1}^\infty\frac{1}{(2n+1)(2n+2)}\\ & = \sum_{n=1}^\infty\left(\frac{1}{(n+1)(n+2)}-\frac{1}{(2n+1)(2n+2)}\right)\\ & =3\sum_{n=1}^{\infty} \frac{n}{(n+2)(2n+1)(2n+2)},\end{align} and this last series obviously converges by the comparison test.
I am able to represent this divergent sum as the difference of two convergent sums. Where is my flaw?
Let $s_m =\sum_{n=1}^{} (-1)^{n}\frac{n}{n+1} $.
Then
$\begin{array}\\ s_{2m} &=\sum_{n=1}^{2m} (-1)^{n}\frac{n}{n+1}\\ &=\sum_{n=1}^{m} ((-1)^{2n-1}\frac{2n-1}{2n-1+1}+(-1)^{2n}\frac{2n}{2n+1})\\ &=\sum_{n=1}^{m} (-\frac{2n-1}{2n}+\frac{2n}{2n+1})\\ &=\sum_{n=1}^{m} \frac{-(2n-1)(2n+1)+4n^2}{2n(2n+1)}\\ &=\sum_{n=1}^{m} \frac{-(4n^2-1)+4n^2}{2n(2n+1)}\\ &=\sum_{n=1}^{m} \frac{1}{2n(2n+1)}\\ \text{and}\\ s_{2m+1} &=s_{2m}+(-1)^{2m+1}\frac{2m+1}{2m+2}\\ &=s_{2m}-\frac{2m+1}{2m+2}\\ &=s_{2m}-(1-\frac{1}{2m+2})\\ \end{array} $
Therefore the even terms converge to $\sum_{n=1}^{\infty} \frac{1}{2n(2n+1)} $ and the odd terms converge to $-1+\sum_{n=1}^{\infty} \frac{1}{2n(2n+1)} $.
Therefore the series does not converge.