Can someone check my explanation of sequences and series?

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

Let S_n 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?

 

[mathman]

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

 

[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 6^th 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 6^th term of the sequence.
The same goes for geometric sequences. Consider
2, 4, 8, 16, 32,...
If I wanted to generate the 6^th 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.

 

[disregardthat]

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}

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.

 

[AMenendez]

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.

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<br /> \\<br /> a_1 = a_0 + d<br /> \\<br /> a_2 = a_1 + d = (a_0 + d) + d = a_0 + 2d<br /> \\<br /> a_3 = a_2 + d = (a_0 + 2d) + d = a_0 + 3d<br /> \\<br /> a_4 = a_3 + d = (a_0 + 3d) + d = a_0 + 4d<br /> \\<br /> \vdots<br /> \\<br /> a_n = a_{n-1} + d = (a_0 + (n-1)d) \rightarrow a_n = a_0 + d(n-1)<br />
Furthermore, for a geometric sequence:
g_0 = g_0<br /> \\<br /> g_1 = g_0 \cdot r<br /> \\<br /> g_2 = g_1 \cdot r = (g_0 \cdot r) \cdot r = g_0 \cdot r^2<br /> \\<br /> g_3 = g_2 \cdot r = (g_0 \cdot r^2) \cdot r = g_0 \cdot r^3<br /> \\<br /> \vdots<br /> \\<br /> g_n = g_0 \cdot r^{n-1}<br />
QED
Keep in mind that a_n or g_n is the n + 1^{th} term.

 

[disregardthat]

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