Series: terminology & notation

[Rasalhague]

(1) Given a sequence

$$\left \langle a_n \right \rangle:\mathbb{N} \rightarrow \mathbb{R} = (a_1,a_2,a_3,...) \; \bigg| \; \left \langle a_n \right \rangle(p) = a_p$$

and another sequence that's the series,

$$\left \langle s_n \right \rangle:\mathbb{N} \rightarrow \mathbb{R} = (s_1,s_2,s_3,...) \; \bigg| \; \left \langle s_n \right \rangle(p) = s_p = \sum_{k=1}^{p} a_p,$$

is there a standard name for the "raw" sequence $\left \langle a_n \right \rangle$ from which $\left \langle s_n \right \rangle$ is constructed: the ... of the series? (The sequence $\left \langle a_n \right \rangle$ of which the series $\left \langle s_n \right \rangle$ is the "sequence of partial sums".)

(2) Have I understood this notation right?

$$\sum_k a_k := \left \langle s_n \right \rangle \;$$

in the context of series. (When I've met this notation before, it's just been a casual notation for a sum, where the codomain of the family (indexing function) is assumed to be known by the reader.) And

$$\sum_{k=1}^{\infty} a_k := \lim_{n \rightarrow \infty } \sum_{k=1}^{n} a_k = \lim_{n \rightarrow \infty } \left \langle s_n \right \rangle ,$$

that is, the sequential limit.

(3) Binmore suggests that $s_n \rightarrow s$ as $k \rightarrow \infty$ implies $s_{n-1} \rightarrow s$ as $k \rightarrow \infty$ (Mathematical Analysis, § 6.9). What exactly does the notation $s_{n-1}$ mean in this context: $\left \langle s_{n-1} \right \rangle (p) = \left \langle s_n \right \rangle (p-1)$, so that $\left \langle s_{n-1} \right \rangle = (?,s_1,s_2,...)$? Or $\left \langle s_{n-1} \right \rangle (p) = \left \langle s_n \right \rangle (p+1)$, so that $\left \langle s_{n-1} \right \rangle = (s_2,s_3,s_4,...) \; ?$ I'm guessing the latter, but, unless I've missed something, he doesn't explicitly define it.

 

[Rasalhague]

(3) Binmore suggests that $s_n \rightarrow s$ as $k \rightarrow \infty$ implies $s_{n-1} \rightarrow s$ as $k \rightarrow \infty$ (Mathematical Analysis, § 6.9). What exactly does the notation $s_{n-1}$ mean in this context: $\left \langle s_{n-1} \right \rangle (p) = \left \langle s_n \right \rangle (p-1)$, so that $\left \langle s_{n-1} \right \rangle = (?,s_1,s_2,...)$? Or $\left \langle s_{n-1} \right \rangle (p) = \left \langle s_n \right \rangle (p+1)$, so that $\left \langle s_{n-1} \right \rangle = (s_2,s_3,s_4,...) \; ?$ I'm guessing the latter, but, unless I've missed something, he doesn't explicitly define it.

Second guess: It seems more natural to read the notation as

$$\left \langle t_n \right \rangle = \left \langle s_{n-1} \right \rangle = \left \langle s_n \right \rangle (p-1) = (?,s_1,s_2,...)$$

so that when $p \neq 1$, we have

$$t_p = s_{p-1}.$$

Although Binmore's notation, if this is what it means, doesn't explicitly define the first entry, $t_1$, I think it must be $t_1 = 0$, because each subsequent entry of $\left \langle t_n \right \rangle$ will include $t_1$ in its sum, and whatever this number is, adding it has no effect, so it must be 0.

 

[Rasalhague]

My last guess seems to work for this example, but is it how the notation is used in general?

When Binmore writes $s_{2^n}$, on p. 56, I think he means a nonspecific entry $s_p$ of the series $\left \langle s_n \right \rangle$ such that $\exists n \in \mathbb{N} (p = 2^n)$. This might suggests that $\left \langle s_{n-1} \right \rangle$ would mean the series

$$\left \langle t_n \right \rangle = (s_1,s_2,s_3,...) = \left \langle s_n \right \rangle$$

since for every entry $s_p$ of the original series, $\exists n \in \mathbb{N} (p = n-1)$. But surely that can't be what he has in mind on p. 57, as it would make a trivial statement.

 

[Office_Shredder]

Basically, you're just right. The sequence s_n-1 is sort of ill-defined at n=1. But this is OK, because we're only interested in what the limit looks like. The claim that $$\lim_{n\to \infty} s_n = \lim_{n\to \infty} s_{n-1}$$ is in fact trivial, given epsilon, if n>N means that |s_n-s|<epsilon, then when n>N+1, |s_n-1-s|<epsilon

 

[samii]

Given the series:

1/1 + 1/(1+2) + 1/(1+2+3) + 1/(1+2+3+4) + ...

How would u find the sum of the series?
Someone please help! :(

 

[Rasalhague]

Thanks, Office Shredder. I've just noticed Binmore does have give some clues about the notation after all, in the chapter on subsequences:

$$\left \langle x_{r+1} \right \rangle = (x_2,x_3,x_4,...)$$

$$\left \langle x_{2r} \right \rangle = (x_2,x_4,x_6,...)$$

and so forth. So we could write

$$\left \langle x_{r+1} \right \rangle (n) := \left \langle x_r \right \rangle (n+1)$$

$$\left \langle x_{2r} \right \rangle (n) := \left \langle x_r \right \rangle (2n)$$

and, in general,

$$\left \langle x_{f(r)} \right \rangle (n) := \left \langle x_r \right \rangle (f(n))$$

where $f:\mathbb{N} \rightarrow \mathbb{N}$. And, in general, $\left \langle x_{r-1} \right \rangle$ would be ill-defined at $n = 1$, unless, as in this case of a series derived from a specified sequence, the definition of the first entry (here the first partial sum) is implicit.

 

[Rasalhague]

Given the series:

1/1 + 1/(1+2) + 1/(1+2+3) + 1/(1+2+3+4) + ...

How would u find the sum of the series?
Someone please help! :(

Hi Samii,

You might have got the answer already, but if not... Each term in any partial sum from this series can be written

$$\frac{1}{\frac{1}{2}(n^2+n)} = \frac{2}{n(n+1)}.$$

For example, where $n=4$,

$$\frac{1}{1+2+3+4} = \frac{2}{n(n+1)} = \frac{2}{20} = \frac{1}{10}.$$

To see why 1+2+3+... = (n/2)(n+1), look at the case where $n$ is an even number, e.g. 4:

1 @
2 @@
3 * * *
4 * * * *

Add the first number to the last. Add the second number to the second to last. And divide by 2:

3+2 * * * @@ = 5
4+1 * * * * @ = 5

This gives

$$\frac{n}{2}(n+1).$$

If you haven't already seen this, try to find a similar formula for odd numbers. With a bit of algebraic manipulation, you'll see that it reduces to the same formula as for even numbers.

If you want to try the next stage yourself, LOOK AWAY NOW!

Otherwise, the sum of the series is then given by

$$\sum_{n=1}^{\infty}\frac{2}{n(n+1)} = 2 \sum_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{n+1} \right )$$

$$= 2 \; \lim \left [ \left ( 1-\frac{1}{2} \right )+\left ( \frac{1}{2}-\frac{1}{3} \right )+\left ( \frac{1}{3}+\frac{1}{4} \right )+... \right ] = ...$$

You can check it at Wolfram Alpha by entering sum (2/(n(n+1))).


But how did I actually go about solving that?

(1) First instinct, cheat! (Actually if I'd had any idea how to solve it, I'd have tried that first, but being stuck, I thought I'd take a short cut.) I entered the series into Mathematica as Sum[1/Sum[k,{k,1,i}],{i,1,Infinity}]. This gave me the answer, and reassured me that an answer did exist. Now I just had to prove it.

(2) Plan A. I skimmed through the relevant chapters of the book I mentioned, Binmore: Mathematical Analysis, in search of a general method for finding the sum of a series. I also looked through the notes I've been making while watching Edward Su's series of lectures on Real Analysis (on YouTube). I found a lot of material on proving whether a sequence has a limit (and hence whether a series has a sum), but I didn't find a general method for actually finding the limit, if there is one. What I did find were proofs establishing the sum of certain specific series. Still in search of general methods, I googled FIND SERIES LIMIT. After a cursory search, I found a few explanations of how to find the limit of specific series, but still no general method.

(3) Plan B. I went back to the examples in the book, hoping one would be similar to this problem. My first thought was to look for some way of showing that the sum of this series is $s \leq 2$, and a proof that $s \geq 2$, which together would show that $s = 2$. In fact, I did find a way to show the former, but the next example I came across gave (effectively) the exact answer, so I just used that instead. Actually, the example in the book showed that

$$\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = 1.$$

The denominator looked familiar, so I compared it to your series and made the nessessary modifications.