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

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


  • $\begingroup$ Make it telescopic. Notice that $\frac{1}{i^2-1}=\frac{1}{2}(\frac{1}{i-1}-\frac{1}{i+1})$. You can achieve this by looking for an expression of the form $\frac{1}{i^2-1}=\frac{A}{i-1}+\frac{B}{i+1}$ and then determining $A$ and $B$ by plugging in different values for $x$. Then add and subtract $\frac{1}{i}$ to make it telescopic to calculate the sum. For the second one, factor out $n^3$ in the numerator and the denominator. $\endgroup$ – stressed out Nov 30 '17 at 3:17

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:


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.

  • $\begingroup$ For the first one, please let me know the full derivation. i tried but failed. For the second one after factoring out n3, i get $\frac{\frac{4}{n} - 1}{ \frac{10}{n^3} +2}$ and then as $\lim_{n\to\infty} $ the result is $-\frac{1}{2} $? $\endgroup$ – user3352074 Dec 2 '17 at 11:30
  • $\begingroup$ Hi, Please show me the full derivation of the first one.Thanks $\endgroup$ – user3352074 Dec 2 '17 at 11:37
  • $\begingroup$ @user3352074: Yes. You're right on the second one. I added more explanation on how to obtain the first one. Don't hesitate to ask for more explanations if you needed. $\endgroup$ – stressed out Dec 2 '17 at 11:47

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