# Partial Fraction Decomposition - "Telescoping sum"

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• opus
In summary: So what you're saying is that for these types of problems, we should be looking for a pattern or a way to simplify the sum using a mathematical tool?Yes, in general, when trying to find a closed form formula for a series, you should look for patterns or try to simplify the sum using known mathematical tools. In this specific problem, partial fraction decomposition was the tool that helped simplify the sum.
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.

Last edited:
opus
opus said:
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
opus said:
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
fresh_42 said:
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?

Mark44 said:
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?

fresh_42 said:
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.

opus said:
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
opus said:
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.

opus said:
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.

opus said:
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."

## What is partial fraction decomposition?

Partial fraction decomposition is a method used to break down a rational expression into simpler fractions. It is useful in solving integration problems and simplifying complex algebraic expressions.

## What is a telescoping sum?

A telescoping sum is a series in which most of the terms cancel out, leaving only a few terms at the beginning and end. This makes it easier to calculate the sum of the series.

## How is partial fraction decomposition used in telescoping sums?

Partial fraction decomposition is used to rewrite a rational expression in the form of a telescoping sum. This allows us to easily identify the terms that cancel out and simplify the sum.

## What is the general process for partial fraction decomposition?

The general process for partial fraction decomposition involves factoring the denominator of the rational expression, setting up equations for the unknown coefficients of the simpler fractions, and solving for these coefficients. The rational expression can then be rewritten as a sum of these simpler fractions.

## When is partial fraction decomposition most useful?

Partial fraction decomposition is most useful when solving integration problems involving rational expressions. It can also be helpful in simplifying complex algebraic expressions, especially when dealing with fractions.

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