The exam exercises for this chapter were much trickier than the homework exercises. With the right approach, they could be solved quickly; too bad I never found that right approach…

## Exam Exercises

### $\sum_{k=1}^n(-1)^k\frac{k}{4k^2-1}$

I managed to solve this exercise using sum by parts; the book uses partial fractions, but as far as I can tell I never learned about these. They seem to be part of high school curriculum in America, but might not be in Belgium (or I was sleeping that day).

Sum by parts worked for me because I eventually asked myself what kind of function of $k$ would produce a finite difference $\frac{1}{4k^2-1} = \frac{1}{(2k+1)(2k-1)}$.

By looking at $\Delta (2k)^{\underline m}$, I realised that the solution was fairly simple. I first experimented with $\Delta (2k)^{\underline{-1}}$, then found the right expression:

For the sum by parts, I can therefore try to use:

The last expression was computed using the product rule for finite difference.

When I put everything into the sum by parts formula, the various blocks felt into place with satisfying “clicks”:

The answer as a function of $n$ is

### 1050

I could not do this exercise; I also failed to see the basic sum that was at the centre of the question (quite prominently).

Even the book solution took me a while to figure out.

The book solution asks how many pairs $a$, $b$ are there such that $\sum_{a\le k \lt b}k = 1050$.

Rewriting this in terms of finite calculus is simple enough:

Now, one thing to notice is that if a sum of two integers is even, so is the difference, and vice-versa. Therefore the product above is the product of one even and one odd integers.

So we are now looking for ways to express

as a product with $x$ is even and $y$ odd.

To compute how many ways there are to produce a divisor of a number whose prime factors are known, it is enough to see that, for each prime factor $p$ with multiplicity $n_p$, this prime can be left out, or included up to $n_p$ times in the divisor; so for each prime p there are $n_p+1$ possibilities. Summing these over the prime factors give the number of divisors of the original number.

In the present case, this number is 12.

Now that we have the number of possible pairs of $x$ and $y$, we have to go back to $a$ and $b$. There might be a principle involved here, but I don’t really see it, so I’ll just try to rebuild the solution from the ground up.

We already know that either $b-a=x$ and $b+a-1=y$ or $b-a=y$ and $b+a-1=x$. So we already see that whatever expression we need, both $a$ and $b$ will have a $\frac{1}{2}$ added so that their difference cancels out, and their sum produces a $1$ that will cancel the $-1$.

Looking at sum and differences, we have $(x+y)-(x-y) = 2y$ and $(x+y)+(x-y)=2x$, so there will be a sum and a difference involved. We also need $a$ to be smaller than $b$, so it is a candidate for $\frac{1}{2}(x-y)+\frac{1}{2}$. However, we also need it to be positive, so we add an absolute value.

So the candidate solutions are $a=\frac{1}{2}|x-y|+\frac{1}{2}$ and $b=\frac{1}{2}(x+y)+\frac{1}{2}$. Let’s check them:

So it all adds up.

### Riemann’s zeta function

This exercise was more in line with the content of this chapter, and easy enough. Essentially, it is just a matter of changing the order of summation.

#### $\sum_{k\ge 2}(\zeta(k)-1) = 1$

The inner sum is a geometric progression:

So we need to solve $\sum_{j\ge 2}\frac{1}{j(j-1)}=\sum_{j\ge 0}\frac{1}{(j+1)(j+2)}$, which we already saw as well, and the value is indeed 1.

#### $\sum_{k\ge 1}(\zeta(2k)-1)$

Using the same approach:

Once again, we have a geometric progression, just as easy to solve as the previous one:

Now we have $\sum_{j\ge 2} j(j-2)^{\underline{-3}}$, which can be summed by parts. I chose:

and now the sum by part:

### $\sum_{k\ge 0}\min(k, x \dot{-} k)$

I am not really sure of my solution here, despite the fact that the outcome is identical to the book; my method is somewhat different from the book’s, and using some concepts from Chapter 3.

To prove that the two sums have the same value, I just evaluate each.

First, a basic observation on the $\dot{-}$ operator: if $b\le 0$, $a\dot{-}b$ is at least zero, and if $a\le 0$, at most $a$ (otherwise it is always zero).

So if $a\le 0$ or $a\le b$, $a\dot{-}b=0$.

I will now assume that $x\ge 0$; otherwise both sums are zero.

### $\sum_{k\ge 0}\min(k,x\dot{-}k)$

First, I replace the infinity sum by a finite one: as seen above, if $k\gt x$, $\min(k,x\dot{-}k)=0$, so

I then try to remove the $\min$ operator:

so that

The general idea here is to find a way to eliminate the $k$ terms, by shifting each term in the second sum by an equal amount.

At this point, the number of terms in each sum is important: the total number of terms is the number of $k$ such that $0\le k\le x$; clearly this is $\lfloor x \rfloor + 1$.

The number of terms in the first sum is similarly $\lfloor\frac{x}{2}\rfloor + 1$.

The number of terms in the second sum is therefore $\lfloor x \rfloor - \lfloor \frac{x}{2} \rfloor$. To find an expression for this, it helps to look at integral $x$ first.

If $x=4$, $\lfloor 4 \rfloor - \lfloor \frac{4}{2} \rfloor = 2$.

If $x=5$, $\lfloor 5 \rfloor - \lfloor \frac{5}{2} \rfloor = 5-2 = 3$.

More generally, if $2n\le \lt 2n+1$, there will be $n+1$ terms in the first sum, and $n$ in the second. And if $2n-1\le x\le 2n$, there will be $n$ terms in each sum.

A first attempt for the number of terms in the second sum is $\lfloor \frac{x}{2} \rfloor$, but this only works for $x$ such that $2n\le x\lt 2n+1$. But it is easy to see that $\lfloor \frac{x+1}{2} \rfloor$ will always work: if $2n\le x\lt 2n+1$, $\lfloor \frac{x+1}{2} \rfloor = \lfloor\frac{2n+1}{2}\rfloor = n$, and if $2n-1\le x\lt 2n$, $\lfloor \frac{x+1}{2}\rfloor = \lfloor\frac{2n}{2}\rfloor = n$.

Using this value to “shift” the terms of the second sum:

Now the question is whether the new $k$ in the second sum cancel the $k$ in the first sum. Once again, let’s check the cases:

- if $2n\le x\lt 2n+1$, $\frac{x}{2} \gt \lfloor\frac{x+1}{2}\rfloor$, so $k\ge 1$: this means the $\lfloor \frac{x}{2}\rfloor$ non-zero $k$ terms of the first sum are cancelled by the $\lfloor \frac{x+1}{2}\rfloor = \lfloor \frac{x}{2}\rfloor$ non zero $k$ terms of the second sum; the zero term can safely be ignored.
- if $2n-1\le x\lt 2n$, $\frac{x}{2} \lt \lfloor\frac{x+1}{2}\rfloor$, so $k\ge 0$; this means the $\lfloor \frac{x}{2}\rfloor+1$ $k$ terms (including 0) are all cancelled by the $\lfloor \frac{x+1}{2}\rfloor = \lfloor \frac{x}{2}\rfloor+1$ $k$ terms of the second sum.

So we can safely rewrite the sum as

And, as we already know the number of terms is $\left\lfloor \frac{x+1}{2}\right\rfloor$, the sum value is

#### $\sum_{k\le 0}(x\dot{-}(2k+1))$

This sum is much easier than the previous one. First I remove the $\dot{-}$ operator. I need $2k+1\le x$:

This gives me $\lfloor \frac{x+1}{2}\rfloor$ number of terms.

So I can extract $x$ and work only on $k$

So the both expressions have the same value.

## Bonus Questions

### $\vee$ Laws

I find it easier to work from the basic $\min$ operator, and scale that up to $\vee$ (the formulas can always be derived mechanically from the underlying algebra).

$\min$ is an associative and commutative operator, is distributive with addition, and its neutral element is $\infty$.

I will not repeat the full list of formulas; the book has them already.

### Undefined infinite sums

An undefined sum, according to (2.59), is one in which both the positive sum and the negative sums are unbounded.

I define $K^+$ as $\{k\in K| a_k \gt 0\}$ and $K^-$ as $\{k\in K| a_k\lt 0\}$.

The point about unbounded sums is that, even if I drop a large number of terms, there are always enough remaining terms to add up to an arbitrary amount.

For instance, given $n$ even and $E_n=K^+\setminus F_{n-1}$, it is always true that I can find $E’_n\subset E_n$ such that

So if I define $F_n = F_{n-1} \cup E’_n$, $\sum_{k\in F_n}a_k \ge A^+$.

And when $n$ is odd, with $O_n=K^-\setminus F_{n-1}$, I can always find a subset $O’_n\subset O_n$ such that

(with $K^-$, the $a_k$ are smaller than zero, so the sum can be arbitrarily small).

If I define $F_n = F_{n-1} \cup O’_n$, $\sum_{k\in F_n}a_k \le A^-$.

As I could not do the other bonus questions, this completes Chapter 2.