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Infinite series?

  • Thread starter pivoxa15
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  • #1
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Homework Statement



IS=infinite series from n=1 to infinity

Does this IS(((-1)^n)/5) converge? But if it did what would it converge to? There is certainly nothing infinite about this sum value but if it dosen't converge to any specific number than does it mean it is divergent?


I think yes.
 

Answers and Replies

  • #2
HallsofIvy
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When in doubt, go back to the definition! The sum of an infinite series is defined as the limit of the sequence of its partial sums, if that sequence converges. An infinite series converges if and only if its sequence of partial sums converges.

In this case the sum is -1/5+ 1/5- 1/5+ 1/5+ ... The partial sums are -1/5, -1/5+ 1/5= 0, -1/5+ 1/5- 1/5= -1/5, -1/5+ 1/5- 1/5+ 1/5= 0, etc. In other words, the sequence of partial sums is -1/5, 0, -1/5, 0, -1/5, 0, ... While that sequence does not go to infinity, it is "oscillating" and still does not converge. Therefore, the infinite sum does not converge.

Neither a sequence nor a series has to "go to infinity" in order to be divergent.
 
  • #3
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Is this the series you want to test: [tex]\sum_{n=1}^{\infty}{\frac{(-1)^n}{5}}[/tex]
?

If so, I think it would depend on how you interpret the sum -1+1-1+1-1...-1+1-1....
 
  • #4
HallsofIvy
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Is this the series you want to test: [tex]\sum_{n=1}^{\infty}{\frac{(-1)^n}{5}}[/tex]
?

If so, I think it would depend on how you interpret the sum -1+1-1+1-1...-1+1-1....
Fortunately the definition of "infinite series" tells us how to interpret it!
 
  • #5
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RIght so now what about [tex]\sum_{n=1}^{\infty}{\frac{(-1)^n}{5+\frac{3}{n}}}[/tex]

When n is large we get back to [tex]\sum_{n=1}^{\infty}{\frac{(-1)^n}{5}}[/tex] which is divergent so the above series is divergent. But is this reason good enough to extablish the divergence of this series?
 
  • #6
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Fortunately the definition of "infinite series" tells us how to interpret it!
Sorry. Major brain lapse. :blushing:
 
  • #7
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RIght so now what about [tex]\sum_{n=1}^{\infty}{\frac{(-1)^n}{5+\frac{3}{n}}}[/tex]

When n is large we get back to [tex]\sum_{n=1}^{\infty}{\frac{(-1)^n}{5}}[/tex] which is divergent so the above series is divergent. But is this reason good enough to extablish the divergence of this series?
Well, does the nth term of the series go to 0 as n goes to infinity?
 
  • #8
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Well, does the nth term of the series go to 0 as n goes to infinity?
The nth term when n goes to infinity is not 0 which is sufficent for the series to diverge.
 
  • #9
Gib Z
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I think Mystic998 Knew that pivoxa15, but wanted to actually know if it actually does go to zero.

Mystic998 - As you can see, there are only 2 values the terms can take, -1/5 and 1/5. Neither of these are zero, so even though we haven't calculated the limit, we know it isn't zero :)
 
  • #10
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There's a different type of "convergence" when you have a series that does that, because the series value will oscillate around some fixed "average", but in the most common definition of convergence, said series doesn't converge.
 
  • #11
HallsofIvy
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I think Mystic988 knew perfectly well that it didn't go to 0 and suggesting a way for pivoxa15 to determine that the series didn't converge!
 
  • #12
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The fact that Mystic988's sentence started with 'Well...' implies he/she knew the answer and wanted to give me a clue by asking me a question. I took it as a clue and found the correct answer as posted in post 8.
 
  • #13
Gib Z
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>.< Sorry I didn't realise it was pivoxa15's question ..I thought he was just another poster trying to help, my bad.
 

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