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Do Alternating Series Have Limits? 
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#1
Apr2812, 04:41 PM

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I would imagine that an alternating series that goes of to infinity doesn't have a limit because it keeps switching back and forth, but I can't find anything in my textbook about it. I just want to make sure that this is right.



#2
Apr2812, 04:48 PM

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Some alternating series do have limits. For example:
[tex]\sum_{n=1}^{+\infty} \frac{(1)^n}{n}[/tex] has a limit (and it equals log(2)). Other series, like [tex]\sum_{n=1}^{+\infty} (1)^n[/tex] have a too large oscillation to have a limit. 


#3
Apr2812, 04:51 PM

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#5
Apr2812, 05:13 PM

P: 12

I'm still kinda confussed, so for example if I have the series:
[tex]\sum_{n=1}^{+\infty} (1)^n \frac{n}{n+1}[/tex] Is the limit nonexistant or does it equal 1? 


#6
Apr2812, 05:14 PM

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[tex]\lim_{n\rightarrow +\infty}{ \frac{(1)^n n}{n+1}}\neq 0[/tex] 


#7
Apr2812, 05:19 PM

P: 12

Okay. So in the previos example where the limit was equal ln2 , do I have to solve for and indeterminate power to get that answer?



#8
Apr2812, 05:21 PM

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#9
Apr3012, 11:07 AM

P: 261

In general , an alternating series of the form [itex]\sum[/itex] (1)^{k} a_{k} will converge if a_{k} [itex]\rightarrow[/itex]0 as k[itex]\rightarrow[/itex]∞ and 0<a_{k+1}≤ a_{k}



#10
Apr3012, 09:19 PM

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If [itex]a_n[/itex] goes to 0 then the series [itex]\sum (1)^na_n[/itex] converges.
If it does not, then we can determine two subseries of [itex]\sum (1)^n a_n[itex], one with n even, the other with n odd, that converge to two different limits. And so the series itself does not converge. 


#11
Apr3012, 09:53 PM

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#12
Apr3012, 09:57 PM

P: 261

Consider [itex]\sum[/itex]_{k=1} (1)^{k+1} a_{k} where a_{k}= 1/k if k is odd and 1/k^{2} if k is even. It is possible to show that this alternating series diverge to +∞ although a_{k} goes to zero. This counter example indicates the necessity of the condition 0<a_{k+1} ≤ a_{k} for convergence to happens. 


#13
Apr3012, 10:36 PM

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If you're asking about the limit of the terms, then the limit does not exist. $$\lim_{n \to \infty} (1)^n \frac{n}{n + 1}\text{ does not exist}$$ The reason is that for large n, successive terms oscillate between values close to 1 and 1, depending on whether n is even or odd, which affects the sign of (1)^{n}. If you're asking about the sum of the series, then there too the limit does not exist. The N^{th} Term Test for Divergence says that if the limit of the terms of the series is different from zero or doesn't exist, then the series diverges. Since I established that the limit of the terms of the series doesn't exist, this theorem says that the series diverges (does not converge). 


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