How to find if a series is telescopic The series
$$\sum_{n=1}^\infty\left(\frac{4n+4}{3n+1}-\frac{4n}{3n-2}\right)$$
is telescopic and it converges to $-4+\dfrac43$.
But if we get the equivalent expresion
$$\sum_{n=1}^\infty\frac{-8}{9n^2-3n-2}$$
Is there an easy criterion to see that it is a telescopic series, or must we hace to express this algebraic fraction as a sum of partial ones? (and cross fingers)
 A: Yes, you try to write your sum as
$$\sum_{n=1}^{+\infty}(A_{n+1}-A_n)$$
So,
$$A_{n+1}=\frac{4n+4}{3n+1}=\frac{4(n+1)}{3(n+1)-2}$$
and, then
$$A_n=\frac{4n}{3n-2}$$
the result is
$$(\lim_{n\to+\infty}A_{n+1})-A_1=\frac 43-4$$
For partial fractions decomposition
$$9n^2-3n-2=(3n+1)(3n-2)$$
and
$$\frac{-8}{9n^2-3n-2}=\frac{a}{3n+1}+\frac{b}{3n-2}$$
with
$$a=-b=\frac 83$$
So, your sum becomes
$$\frac 83\sum_{n=1}^{+\infty}(B_{n+1}-B_n)=$$
$$\frac 83(\lim_{n\to+\infty}B_{n+1}-B_1)=$$
$$\frac 83(0-1)=-\frac 83=\frac 43-4$$
A: $B_n=\frac{4n+4}{3n+1}$ and $C_n=\frac{4n}{3n-2}$ with $A_n=B_n-C_n$.  However $B_{n-1}=C_n$, so the series collapses.
A: In fact, every series can be written as a telescopic, by writing its terms as the difference between the successive partial sums:
$$\sum_{n=1}^\infty a_n = \sum_{n=1}^\infty (s_n-s_{n-1})$$
where $s_n=a_1+a_2+...+a_n$ are the partial sums and say, by convention, $s_0=0$. So, what is needed in cases such as the OP, is to investigate whether there is some useful transformation of the original terms $a_n$ into a telescopic form, as a way of determining convergence/divergence (as done in hamam_Abdallah's solution).
