General formula for the partial sums I m having trouble figuring out how to find the general formula for partial sums of the following two series.


*

*$\sum_{i=2}^n \frac{1}{i^2-1} = \frac{3}{4}-\frac{1}{2n}-\frac{1}{2(n+1)}$

*$\lim_{n\to\infty} \frac{4n^2 -n^3}{10+2n^3} = -\frac{1}{2} $


Both are in convergence and divergence section in the link at example-2 and example-5 respectively.
Let me know how those are being derived
thanks
 A: Make it telescopic. For the first one, notice that:$$\frac{1}{i^2-1}=\frac{1}{2}(\frac{1}{i-1}-\frac{1}{i}+\frac{1}{i}-\frac{1}{i+1})$$ 
This is obtained by noting that $i^2-1$ in the denominator can be factored into two linear terms, namely $i+1$ and $i-1$. Then we assume we have a factorization in the form of:
$$\frac{1}{i^2-1}=\frac{A}{i-1}+\frac{B}{i+1}$$
This gives:
$$A(i+1)+B(i-1)=1$$
which has to hold for any value of $i$. Plug in $i=-1$ and $i=+1$ to see that $A=\frac{1}{2}$ and $B=-\frac{1}{2}$.
This gives 
$$\frac{1}{i^2-1}=\frac{1}{2}\left(\frac{1}{i-1}-\frac{1}{i+1}\right)$$
Now this looks very close to our goal. We are just one step away. Add and subtract $\frac{1}{i}$ and you will obtain the desired telescopic form.
For the second one, factor out the highest power and use what you know about limits when $n$ goes to infinity. This is how it's done:
$$\lim_{n\to \infty}\frac{4n^2-n^3}{10+2n^3}=\lim_{n\to \infty}\frac{n^3(\frac{4}{n}-1)}{n^3(\frac{10}{n^3}+2)}=\frac{-1}{2}$$
Note that $n^3$ in the numerator and the denominator are cancelled and you are left with something that goes to zero except for the constant terms. 
