PDA

View Full Version : A Stupid(?) Question

RocketManJames
08-25-2005, 06:16 PM
Sum of Geometric Series converges when C &lt; 1. One of the proofs of this that I learned back in school follows.

So, S = C^0 + C^1 + C^2 + ...
C*S = C^1 + C^2 + C^3 + ...
S - C*S = C^0
S = C^0 / (1 - C) = 1 / (1 - C)

Now, comes the stupid question... how does this proof fail for C &gt; 1? What is the reason why this proof is valid when C &lt; 1 and invalid when C &gt; 1? At C = 1, we clearly see that the answer is undefined... but at C &gt; 1, S is equal to some negative number, which is obviously wrong. But, my question, lame as it may be, is exactly where is it that this proof becomes invalid?

Thanks.

-RMJ

BruceZ
08-25-2005, 06:22 PM
[ QUOTE ]
Sum of Geometric Series converges when C &lt; 1. One of the proofs of this that I learned back in school follows.

So, S = C^0 + C^1 + C^2 + ...
C*S = C^1 + C^2 + C^3 + ...
S - C*S = C^0
S = C^0 / (1 - C) = 1 / (1 - C)

Now, comes the stupid question... how does this proof fail for C &gt; 1? What is the reason why this proof is valid when C &lt; 1 and invalid when C &gt; 1? At C = 1, we clearly see that the answer is undefined... but at C &gt; 1, S is equal to some negative number, which is obviously wrong. But, my question, lame as it may be, is exactly where is it that this proof becomes invalid?

Thanks.

-RMJ

[/ QUOTE ]

When determining the sum of a series, it is necessary to show that the series converges, not just what the series would be equal to if it did converge. For C &gt; 1, it is easy to show that the series does not converge, or more precisely, that the sequence of partial sums does not converge to a limit. S - C*S = C^0 doesn't make any sense when S and C*S are infinite.

AaronBrown
08-26-2005, 01:49 PM
This is an excellent question. BruceZ's answer is correct, in general you can have a set of perfectly legal steps that leads to an incorrect result if there in an infinity mixed in. But this case also has a more tangible answer. Do the finite case:

S = C^0 + C^1 + C^2 + ... + C^N
C*S = C^1 + C^2 + C^3 + ... + C^(N+1)
S - C*S = [C^0 - C^(N+1)]
S = [1 - C^(N+1)] / (1 - C)

If |C|&lt;1 then C^(N+1) goes to zero as N goes to infinity. So the limit of the finite sum equals 1/(1 - C). Another way of saying that is the series converges to 1/(1 - C). But if |C| &gt; 1 then the sum diverges, there is no finite limit.

BruceZ
08-26-2005, 02:16 PM
[ QUOTE ]
This is an excellent question. BruceZ's answer is correct, in general you can have a set of perfectly legal steps that leads to an incorrect result if there in an infinity mixed in. But this case also has a more tangible answer. Do the finite case:

S = C^0 + C^1 + C^2 + ... + C^N
C*S = C^1 + C^2 + C^3 + ... + C^(N+1)
S - C*S = [C^0 - C^(N+1)]
S = [1 - C^(N+1)] / (1 - C)

If |C|&lt;1 then C^(N+1) goes to zero as N goes to infinity. So the limit of the finite sum equals 1/(1 - C). Another way of saying that is the series converges to 1/(1 - C). But if |C| &gt; 1 then the sum diverges, there is no finite limit.

[/ QUOTE ]

[1 - C^(N+1)] / (1 - C) is the sequence of partial sums which I mentioned. An infinite series is defined as the limit of a sequence of partial sums. This sequence of partial sums has no limit for |C| &gt; 1, hence the series diverges by definition.

This definition brings more clarity to the elementary notion that an infinite series is something of the form x1 + x2 + x3 + .... as it makes clear what the "..." means, and helps prevent us from trying to sum series which don't converge. A similar notion allows us to make sense of such things as continued fractions:

1+1/(1+1/(1+1/(1+...

and continued square roots:

sqrt(1+sqrt(1+sqrt(1 + ...

BTW, both of these equal the golden ratio 1/2 + sqrt(5)/2.