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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)

ajotatxe
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    Often the trick is to do partial fraction decomposition. If you want a telescopic series you need a difference, try to obtain that. There is no rigorous criterion for this, as far as my knowledge goes. Perhaps someone else can help with that? – Ritam_Dasgupta Jun 01 '21 at 21:02
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    Telescopic can only be used if you have a difference inside your infinite sum. As you have a polynomial of degree $2$ in the denominator, you should always try to factor that out and then you could try partial fraction decomposition – LegNaiB Jun 01 '21 at 21:03
  • The usual trick for this sort of problem is to use partial fractions. – Thomas Andrews Jun 01 '21 at 21:06
  • @ThomasAndrews So there's no efficient way to see to what converges $\sum\dfrac1{an^2+bn+c}$ even if $b^2-4ac>0$. – ajotatxe Jun 01 '21 at 21:13
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    No, the partial fraction approach only works for sure when the numerator is a constant and the denominator is of the form $$(ax+b)(a(x+1)+b)$$ @ajotatxe – Thomas Andrews Jun 01 '21 at 21:15

3 Answers3

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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

$$\frac 83\sum_{n=1}^{+\infty}(B_{n+1}-B_n)=$$ $$\frac 83(\lim_{n\to+\infty}B_{n+1}-B_1)=$$ $$\frac 83(0-1)=-\frac 83=\frac 43-4$$

Pat
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hamam_Abdallah
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$B_n=\frac{4n+4}{3n+1}$ and $C_n=\frac{4n}{3n-2}$ with $A_n=B_n-C_n$. However $B_{n-1}=C_n$, so the series collapses.

herb steinberg
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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).

Pat
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