# Partial Fraction Decomposition - "Telescoping sum"

• B
• opus

#### opus

Gold Member
There is a problem in a PreCalculus book that I'm going over that states:

Express the sum ##\frac{1}{2⋅3}+\frac{1}{3⋅4}+\frac{1}{4⋅5}+...+\frac{1}{2019⋅2020}## as a fraction of whole numbers in lowest terms.

It goes on to state that each term in the sum is of the form ##\frac{1}{k\left(k+1\right)}## which is obvious.

The partial fraction decomposition of this is ##\frac{1}{k\left(k+1\right)}=\frac{1}{k}-\frac{1}{k+1}## Again, no problems here.

Now it follows that by using this decomposition for each term, we get:
##\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+...+\left(\frac{1}{2019}-\frac{1}{2020}\right)##

After removing the parenthesis, the only terms that are left, after adding them all up, is the first and last term.

My question is, in being presented with this problem, what "flags" if you will, should we be looking that tells us to use partial fraction decomposition? In trying this problem myself, I got to just the basic setup of ##\frac{1}{k\left(k+1\right)}##, but from here, I would never guess that it needed to be decomposed.

And an additional question- we can clearly see that after the parentheses are removed, that terms start cancelling, but what's to say that 20 terms down the line, they don't cancel? How can we be sure?

That's a matter of experience or persistence. If you try long enough without any efforts, you finally try everything and the one working path will be among them. On the other hand does experience a similar job. There are a handful of tricks which by experience migrate into your standard repertoire and partial fractions are among them.

The usual way, however, is to check small numbers and see if you can recognize a pattern. Latest at ##10## the pattern should be obvious.

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• opus
And an additional question- we can clearly see that after the parentheses are removed, that terms start cancelling, but what's to say that 20 terms down the line, they don't cancel? How can we be sure?
Write the sum with more terms. In the expression below, where you have the ellipsis (...), fill in three or four general terms (terms involving n), and you can see which terms cancel with which other terms.
##\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+...+\left(\frac{1}{2019}-\frac{1}{2020}\right)##

• opus
And an additional question- we can clearly see that after the parentheses are removed, that terms start cancelling, but what's to say that 20 terms down the line, they don't cancel? How can we be sure?
Write it with sums:
$$\sum_{k=2}^{2019}\dfrac{1}{k(k+1)}=\sum_{k=2}^{2019}\left( \dfrac{1}{k}-\dfrac{1}{k+1} \right)=\sum_{k=2}^{2019}\dfrac{1}{k}-\sum_{k=3}^{2020}\dfrac{1}{k}=\sum_{k=2}^{2}\dfrac{1}{k}-\sum_{k=2020}^{2020}\dfrac{1}{k}+\sum_{k=3}^{2019}\left(\dfrac{1}{k}-\dfrac{1}{k} \right)=\dfrac{1}{2}-\dfrac{1}{2020}$$

• opus
That's a matter of experience or persistence. If you try long enough you without any efforts, you finally try everything and the one working path will be among them. On the other hand does experience a similar job. There are a handful of tricks which by experience migrate into your standard repertoire and partial fractions are among them.

The usual way, however, is to check small numbers and see if you can recognize a pattern. Latest at ##10## the pattern should be obvious.

That sounds fair. So this isn't a cover-all case, and there can be similar cases which wouldn't necessarily have these same steps and reasoning?

Write the sum with more terms. In the expression below, where you have the ellipsis (...), fill in three or four general terms (terms involving n), and you can see which terms cancel with which other terms.
##\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+...+\left(\frac{1}{2019}-\frac{1}{2020}\right)##

Ok so once we can start to see a pattern of reasonable length, we can safely assume that the pattern will continue?

Write it with sums:
$$\sum_{k=2}^{2019}\dfrac{1}{k(k+1)}=\sum_{k=2}^{2019}\left( \dfrac{1}{k}-\dfrac{1}{k+1} \right)=\sum_{k=2}^{2019}\dfrac{1}{k}-\sum_{k=3}^{2020}\dfrac{1}{k}=\sum_{k=2}^{2}\dfrac{1}{k}-\sum_{k=2020}^{2020}\dfrac{1}{k}+\sum_{k=3}^{2019}\left(\dfrac{1}{k}-\dfrac{1}{k} \right)=\dfrac{1}{2}-\dfrac{1}{2020}$$

What do those big E symbols mean? Is that Calculus? I haven't been introduced to those yet.

That sounds fair. So this isn't a cover-all case, and there can be similar cases which wouldn't necessarily have these same steps and reasoning?
Yes. After a few dozens of (different) cases, in which partial fractions helped, you automatically get used to check this possibility. Latest if it comes to integration, you should remember partial fraction decomposition.
What do those big E symbols mean? Is that Calculus? I haven't been introduced to those yet.
That is the big Greek letter ##S## which stands for "sum". It's an abbreviation to avoid those dots. It simply means
$$\sum_{i=1}^{N} a_i = a_1 +a_2 + a_3 + a_4 + ... \text{ many } a_i \text{ later} \ldots +a_{N-2} +a_{N-1}+a_N$$
It is a method to handle such sums by administrating the indices instead.

• opus
Ok that makes sense. So it's safe to say that partial fraction decomposition is a very important skill and I'll be using it later?

• fresh_42
Ok that makes sense. So it's safe to say that partial fraction decomposition is a very important skill and I'll be using it later?
Probably. For integration it is an important tool. See in your example: ##\dfrac{1}{k^2+k}## is much more inconvenient than ##\dfrac{1}{k}## or ##\dfrac{1}{k+1}## is.

• opus
Ok great. Thanks guys.

My question is, in being presented with this problem, what "flags" if you will, should we be looking that tells us to use partial fraction decomposition?

The aspect of partial fractions is a special case for some series, however the idea of "telescoping sum" is quite general.

In general, if you want to find a "closed form" formula for the series ##\sum_{i=1}^n f_i = f_1 + f_2 + ... f_n## you try to find a function ##F(k)## such that ##F(k+1) - F(k) = f_k##. This technique is called "summation by anti-differencing".

##\sum_{i=1}^n f_i = f_1 + f_2 + ...f_n = (F(2)-F(1) + (F(3) - F(2)) + ...(F(n+1)-F(n)) ##
## = F(n+1) - F(1)##.

Results like this are studied in the "Calculus of Finite Differences", which is often taught after people have studied calculus, but can be profitably studied before taking calculus.

For example, the result ##\sum_{i=1}^n = 1 + 2 + ...n = n(n+2)/2 ## is often taught by telling how the young Gauss derived the formula, but a more general way is to look for a second degree polynomial ##F(k) = Ak^2 + Bk + C## such that ##F(k+1) - F(k) =k ##. This leads to ##A = 1/2, B= -1/2## and ##C## an arbitrary constant.

That approach can be applied to deriving formulae for sums like ##\sum_{i=1}^n i^3##. In that case, we'd be looking for a fourth degree polynomial.

And an additional question- we can clearly see that after the parentheses are removed, that terms start cancelling, but what's to say that 20 terms down the line, they don't cancel? How can we be sure?
You use the final Peano axiom for "natural numbers": The principle of induction.

From (https://en.wikipedia.org/wiki/Peano_axioms):

"The intuitive notion that each natural number can be obtained by applying successor sufficiently often to zero requires an additional axiom, which is sometimes called the axiom of induction.

1. If K is a set such that:
• 0 is in K, and
• for every natural number n, n being in K implies that S(n) is in K,
then K contains every natural number."