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series Converges. |
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| Aug5-12, 02:28 PM | #1 |
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series Converges.
When I say a series [itex]\sum[/itex]a[itex]_{n}[/itex] converges, what exactly is it that I am saying?
for example [itex]\sum^{∞}_{n=1}[/itex][itex]\frac{9n^{2}}{3n^{5}+5}[/itex] is convergent. what did I just say? |
| Aug5-12, 03:07 PM | #2 |
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Why would you say a series converges if you don't know what it means?
If you have taken a course dealing with sequences and series, then you should have seen a definition of "convergence of a sequence": the series [itex]\sum_{n=1}^\infty a_n[/itex] converges if and only if the sequence of partial sums [itex]s_i= \sum_{n= 0}^i a_n[/itex] converges. (I hope you won't say that [itex]\sum_{n=1}^\infty \frac{9n^2}{3n^2+ 5}[/itex] is convergent. It obviously isn't.) |
| Aug5-12, 03:21 PM | #3 |
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Congrats on telling me exactly what the book told . so now if you don't mind tell it to me as if I was not a person studying Mathematics .
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| Aug5-12, 03:59 PM | #4 |
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Recognitions:
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series Converges. |
| Aug5-12, 04:16 PM | #5 |
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Recognitions:
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A series converges to a value "s" if, by adding enough terms , you can get indefinitely
close to the value s. This is made rigorous : if I want to be within, say, 1/100 of the value s, I must show that there is a term ,say "N", so that by adding N-or-more terms, the value of the expression: (a_1+a_2+......+a_N )-s of the sum will be within 1/100 of the value s. Take the series 1+1/2+1/4+....... Its limit is 2. After 1 term, you are within 1 unit of the limit. After adding two terms you are within 1/2 of the limit. Now, convergence means that I can guarantee that , no matter how close I want to get to 2, I just need to add enough terms, and my sum will be within this --or (almost) any other--distance from 2. We do not demand that the sum be exactly two, but that the sum be indefinitely close to it. |
| Aug5-12, 05:09 PM | #6 |
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Thank you guys very much I now understand.
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