Do Alternating Series Have Limits?

[I'm Awesome]

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.

 

[micromass]

Some alternating series do have limits. For example:

$$\sum_{n=1}^{+\infty} \frac{(-1)^n}{n}$$

has a limit (and it equals log(2)).

Other series, like

$$\sum_{n=1}^{+\infty} (-1)^n$$

have a too large oscillation to have a limit.

 

[Dickfore]

Some alternating series do have limits. For example:

$$\sum_{n=1}^{+\infty} \frac{(-1)^n}{n}$$

has a limit (and it equals log(2)).

Actually $-\ln 2$.

 

[micromass]

Actually $-\ln 2$.

Ah yes, thank you!

 

[I'm Awesome]

I'm still kinda confussed, so for example if I have the series:
$$\sum_{n=1}^{+\infty} (-1)^n \frac{n}{n+1}$$


Is the limit nonexistant or does it equal 1?

 

[micromass]

I'm still kinda confussed, so for example if I have the series:
$$\sum_{n=1}^{+\infty} (-1)^n \frac{n}{n+1}$$


Is the limit nonexistant or does it equal 1?

That limit doesn't exist since

$$\lim_{n\rightarrow +\infty}{ \frac{(-1)^n n}{n+1}}\neq 0$$

 

[I'm Awesome]

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?

 

[micromass]

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.

 

[Useful nucleus]

In general , an alternating series of the form $\sum$ (-1)^k a_k will converge if a_k $\rightarrow$0 as k$\rightarrow$∞ and 0<a_k+1≤ a_k

 

[HallsofIvy]

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

 

[DonAntonio]

If $a_n$ goes to 0 then the series $\sum (-1)^na_n$ converges.



*** The convergence to zero of $a_n$ 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$, one with n even, the other with n odd, that converge to two different limits. And so the series itself does not converge.$[/itex]

[itex]$<br /> ...$[/itex]

 

[Useful nucleus]

If $a_n$ goes to 0 then the series $\sum (-1)^na_n$ converges.

Not necessary for the following:
Consider $\sum$_k=1 (-1)^k+1 a_k where a_k= 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 a_k goes to zero.
This counter example indicates the necessity of the condition 0<a_k+1 ≤ a_k for convergence to happens.

 

[Mark44]

I'm still kinda confussed, so for example if I have the series:
$$\sum_{n=1}^{+\infty} (-1)^n \frac{n}{n+1}$$


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 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).