# Can someone check my explanation of sequences and series?

1. Apr 12, 2014

### Vishera

It's been a while since I've dealt with sequences and series. Here is my explanation of sequences and series and let me know if I am right or wrong.

A sequence is just a list of numbers. By convention, we use the letter $a$ for sequences and they are written in a form like so: $a_1,a_2,a_3,a_4,...$

A sequence can be finite or infinite.

1,2,3,4 is a finite sequence. 1,2,3,4,... is an infinite sequence.

An arithmetic sequence, for some constant d: $a_n=a_0+dn$
A geometric sequence, for some constant r: $a_{ n }=a_{ 0 }r^{ n }$

A series is the sum of the terms of a sequence. By convention, is there a letter for series? I can't remember. Let us use the letter S in the meantime. Series are written like so: $S_1,S_2,S_3,S_4,...$

Let Sn be the series of the finite sequence mentioned earlier. S1=1. S2=3. S3=6. S4=10.

Let Sn be the series of the infinite sequence mentioned earlier. S=∞

I feel like I'm doing something wrong. Can anyone briefly mention which parts are wrong?

2. Apr 12, 2014

### mathman

Except for minor typo (you have Sn when you meant S in last line), looks fine.

3. Apr 13, 2014

### AMenendez

When you explain arithmetic and geometric sequences, the general term expression(s) are given by:
$a_n = a_0 + d(n-1)$
$g_n = g_0 \cdot r^{n-1}$
The $n-1$ is necessary to generate the term; for example, the sequence
$1, 2, 3, 4, 5,...$
is arithmetic, and if I wanted to generate the 6th term of the sequence, I would use:
$a_6 = 1 + 1(6-1) = 6$
Had I used your expression, $a_n = a_0 + dn$, then the term would be:
$a_6 = 1 + 1(6) = 7$, which is obviously not the 6th term of the sequence.
The same goes for geometric sequences. Consider
$2, 4, 8, 16, 32,...$
If I wanted to generate the 6th term of this sequence, I would use:
$g_6 = 2 \cdot (2)^{6-1} = 2 \cdot 32 = 64$
Again, if I used your expression:
$g_6 = 2 \cdot (2)^{6} = 2 \cdot 64 = 128$
So really what your expressions do is generate the $n + 1 ^{th}$ term of the sequence, rather than the $n^{th}$
Your explanations of series are also fine, but I might add the more formal definition of a series:
$S_n = \sum_{i=0}^{n} t_i$ which is really just the sum of some number of terms.Furthermore, an infinite sequence does not always add to infinity, as a geometric sequence with a ratio $r$ such that $-1 < r < 1$ converges as $n \rightarrow \infty$. The example that you gave, though, will diverge to $\infty$.

4. Apr 14, 2014

### disregardthat

Well, this would imply that $a_0 = a_0 -d$, which is not quite right. You probably mean $a_{n-1}$, but there's nothing wrong with Visheras explanation.

Last edited: Apr 14, 2014
5. Apr 14, 2014

### AMenendez

I'm not sure what you're arguing, as $a_{n+1}$ implies a recursion, in which case (for an arithmetic progression) $a_{n+1} = a_n + d$
Consider:
$a_0 = a_0 \\ a_1 = a_0 + d \\ a_2 = a_1 + d = (a_0 + d) + d = a_0 + 2d \\ a_3 = a_2 + d = (a_0 + 2d) + d = a_0 + 3d \\ a_4 = a_3 + d = (a_0 + 3d) + d = a_0 + 4d \\ \vdots \\ a_n = a_{n-1} + d = (a_0 + (n-1)d) \rightarrow a_n = a_0 + d(n-1)$
Furthermore, for a geometric sequence:
$g_0 = g_0 \\ g_1 = g_0 \cdot r \\ g_2 = g_1 \cdot r = (g_0 \cdot r) \cdot r = g_0 \cdot r^2 \\ g_3 = g_2 \cdot r = (g_0 \cdot r^2) \cdot r = g_0 \cdot r^3 \\ \vdots \\ g_n = g_0 \cdot r^{n-1}$
QED
Keep in mind that $a_n$ or $g_n$ is the $n + 1^{th}$ term.

6. Apr 14, 2014

### disregardthat

If you look at the sentence I quoted you will see what I mean.