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I'm reading the resolution of a solved exercise about Telescoping Series in a "Finite Math" course I'm attending. The resolution has a step that is not explained because it's left as an exercise to the reader:

$$ \cdots = \frac{1}{(2k + 3)(2(k + 1) +3)} = \frac{1}{2} \bigg(\frac{1}{2k + 3} - \frac{1}{2(k + 1) + 3}\bigg)$$

I'm guessing the missing step(s) - that turned the fraction into the last "half a difference" of fractions - uses "Partial Fraction Decomposition" technique(s) (that is also part of the Syllabus of this course), but I'm not sure about that and I'm not yet familiar with that technique(s).

So, I read the Wikipedia article about "Partial Fraction Decomposition", namely the "Example 1" section and tried to apply it here, but I get "stuck" quickly:

$$\frac{1}{(2k + 3)(2(k + 1) +3)} = \frac{A}{2k + 3} + \frac{B}{2(k + 1) + 3} \Leftrightarrow$$

$$\Leftrightarrow A[2(k + 1)+3] + B(2k +3) = 1$$

... and I don't know how to carry on from here.

Am I right when I think this is solved using "Partial Fraction Decomposition"? Have I begun solving this well? What should I do now?

ricmarques
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2 Answers2

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You want to cross multiply the fractions and then equate both sides.

We have:

$$1 = A(2k + 5) + B(2k+3)$$

This leads to two equations with two unknowns.

  • $A+B = 0$
  • $5A + 3B = 1$.

Solving these leads to $A = \dfrac{1}{2}, B = -\dfrac{1}{2}$.

You might also want to look up Partial Fraction Cover Up Method and using limits to solve for the constants.

Amzoti
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    Nice work! $\langle +1\rangle$ – amWhy Sep 01 '13 at 18:02
  • @Amzoti Thanks for the answer and for the reference to "Partial Fraction Cover Up Method" :-) I upvoted your reply but ended up accepting the answer of "wfwood" because it was similar, a minute earlier than yours and - last but not least - "wfwood" seems to "need" the extra reputation a bit more than you! ;-) I would like to accept both answers, but I can't :-( – ricmarques Sep 01 '13 at 18:07
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    @ricmarques: You are welcome and no worries, the goal is to help people and I wish we could hide all of our reputation, because the best answers should rise to the top! :-) Regards – Amzoti Sep 01 '13 at 18:08
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This breaks down into
$(A2+B2)k+ (5A+3B) = 1$ or since k is an indeterminate...
$(A2+B2)k+ (5A+3B) = 1 + 0k$ So from there you simply match up the coefficients. $(A2+B2) = 0 $ & $(5A+3B)=1$ You can then solve for A and B.

wfw
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