How to get the expression for this partial sum
I was asked to calculate what the sum of the following series is $$\sum_{n=1}^\infty\frac{2}{4n^2-1}$$ I split up the fraction into $$\frac{1}{2n-1}-\frac{1}{2n+1}$$ From here I wasn't sure what to do so I checked the answers which said that from this we can see that the partial sum up to $m$ is given by $$\sum_{n=1}^{m}\frac{2}{4n^2-1}=1-\frac{1}{2m+1}\tag{1}$$ and that this converges to $1$ as $m\to\infty$. I don't understand how $\frac{1}{2m+1}\to0$ and how they came to the equation in $(1)$.
EDIT: The question on why $\frac{1}{2m+1}\to0$ was in reference to why the series $\sum_{n=1}^{m}\frac{2}{4n^2-1}$ doesn't tend to $0$ because both sequences $\frac{1}{2n-1}$ and $\frac{1}{2n+1}$ tend to zero. I accidently overlooked the difference.
Solution 1:
Note that\begin{align}\require{cancel}\sum_{n=1}^m\left(\frac1{2n-1}-\frac1{2n+1}\right)&=1-\cancel{\frac13}+\cancel{\frac13}-\cdots-\cancel{\frac1{2m-1}}+\cancel{\frac1{2m-1}}-\frac1{2m+1}\\&=1-\frac1{2m+1}.\end{align}