Do Alternating Series Have Limits?

In summary, the limit of the terms and the limit of the series in question both do not exist, making it divergent.
  • #1
I'm Awesome
14
0
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.
 
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  • #2
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
micromass said:
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)).

Actually [itex]-\ln 2[/itex].
 
  • #4
Dickfore said:
Actually [itex]-\ln 2[/itex].

Ah yes, thank you!
 
  • #5
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
I'm Awesome said:
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?

That limit doesn't exist since

[tex]\lim_{n\rightarrow +\infty}{ \frac{(-1)^n n}{n+1}}\neq 0[/tex]
 
  • #7
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
I'm Awesome said:
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?

No. You get that answer if you find the Taylor series for the logarithm.
 
  • #9
In general , an alternating series of the form [itex]\sum[/itex] (-1)k ak will converge if ak [itex]\rightarrow[/itex]0 as k[itex]\rightarrow[/itex]∞ and 0<ak+1≤ ak
 
  • #10
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.
 
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  • #11
HallsofIvy said:
If [itex]a_n[/itex] goes to 0 then the series [itex]\sum (-1)^na_n[/itex] converges.



*** The convergence to zero of [itex]a_n[/itex] must be monotone, otherwise the Leibnitz test may fail.

DonAntonio ***


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.


...
 
  • #12
HallsofIvy said:
If [itex]a_n[/itex] goes to 0 then the series [itex]\sum (-1)^na_n[/itex] converges.
Not necessary for the following:
Consider [itex]\sum[/itex]k=1 (-1)k+1 ak where ak= 1/k if k is odd and 1/k2 if k is even. It is possible to show that this alternating series diverge to +∞ although ak goes to zero.
This counter example indicates the necessity of the condition 0<ak+1 ≤ ak for convergence to happens.
 
  • #13
I'm Awesome said:
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?

First off, it's not clear to me whether you're asking about the limit of the terms in the series or the limit of the series itself.

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 Nth 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).
 

1) Do alternating series always have limits?

Yes, alternating series always have limits. This is because alternating series are a type of infinite series that alternate between positive and negative terms, and as long as the terms are decreasing in magnitude, the series will have a limit.

2) How do you determine the limit of an alternating series?

To determine the limit of an alternating series, you can use the Alternating Series Test, which states that if the terms of an alternating series are decreasing in magnitude and approach 0, then the series will converge to a limit. Additionally, you can use other convergence tests such as the Ratio Test or the Root Test to determine the convergence of an alternating series.

3) Can an alternating series have multiple limits?

No, an alternating series can only have one limit. This is because by definition, an alternating series must alternate between positive and negative terms, and if it has multiple limits, then the terms would not be alternating.

4) Can an alternating series diverge?

Yes, an alternating series can diverge. This can happen if the terms of the series do not decrease in magnitude or if they do not approach 0. In this case, the series will not converge to a limit and will instead diverge.

5) Are there any other types of infinite series besides alternating series?

Yes, there are many other types of infinite series, including geometric series, telescoping series, and power series. Each of these types has its own set of convergence tests and properties that determine whether or not the series converges or diverges.

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