0
$\begingroup$

Without going to much into details about the question itself I would like to draw attention to the fact that Spivak assumes knowledge of a formula (I got it from the solutions in the back of the book) which, in my opinion, he did not present earlier. Once this formula is used the problem becomes trivial.

$$ \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1} $$

Unfortunately I would not be able to arrive at this formula on my own. Therefore, I either missed something in Spivak's text or I am missing a trivial yet essential insight.

Please explain how a mathematics newbie is supposed to know that. Thanks in advance.

$\endgroup$
1
0
$\begingroup$

This (going from LHS to RHS) is called a partial fraction decomposition.

It's easy to "see" when you go from RHS to LHS. Just put everything over a single denominator:

$$\frac 1k - \frac{1}{k+1} = \frac{(1)(k+1) - (1)(k)}{k(k+1)} = \frac{1}{k(k+1)}$$

so you know the equation to be true.

However to go from LHS to RHS, it's slightly more involved.

You can start by letting $\displaystyle \frac{1}{k(k+1)} = \frac{A}{k} + \frac{B}{k+1}$.

$A$ and $B$ are constants.

You can rearrange the RHS to $\displaystyle \frac{A(k+1) + Bk}{k(k+1)} = \frac{(A+B)k + A}{k(k+1)}$.

Now, for the numerator, equate coefficients of the $k$ and the constant terms separately, and solve the simultaneous equations to get the values for $A$ and $B$. This is one basic method of partial fraction decomposition.

There are other shortcuts such as the "Heaviside cover up rule".

You can look up other references to see exactly how partial fractions work. I cannot cover it exhaustively here.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.