Alternating Series: Does Divergence Hold?

[tandoorichicken]

I know that a series such as

\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}

is divergent. Is this also the case for an alternating version of the same series, i.e.,

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

?

 

[dextercioby]

What criteria do you have for alternating series?

Daniel.

 

[Data]

As dexter hinted to, look up the alternating series test. A stronger version is Abel's test, and even stronger is the Dirichlet test.

 

[tandoorichicken]

Well, I know about the alternate series test, I am just saying that is it reliable to say that if a particular series does not converge to a certain sum, then a similar series that alternates between positive and negative also will not converge to a specific sum?

 

[Data]

well, it depends on the series. The alternating series test says that

\sum_{n=0}^\infty (-1)^na_n, \; \mbox{with} \ a_n \geq a_{n+1} \ \forall n \geq 0

converges if \lim_{n \rightarrow \infty} a_n = 0.

So in your specific example, the alternating series converges, even though the series is not absolutely convergent, because

\frac{1}{\sqrt{n+1}} \leq \frac{1}{\sqrt{n}} \ \forall n \geq 1

and

\lim_{n \rightarrow \infty} \frac{1}{\sqrt{n}} = 0

If a series is convergent but not absolutely convergent, we call it conditionally convergent. Conditionally convergent series have some very unintuitive properties, one of which is described here:

http://mathworld.wolfram.com/RiemannSeriesTheorem.html

 

[dextercioby]

The sum is approximately 0.6...


Daniel.

 

[saltydog]

Regarding the MathWorld reference:

Can anyone explain to me how to calculate these sums?

\sum_{k=1}^{\infty}\frac{1}{4k(2k-1)}=\frac{1}{2}ln(2)

\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}=ln(2)

 

[dextercioby]

The last is very easy,if u consider the Taylor series of \ln(1+x) around zero...(there's another elegant construction,too).

As for the first,write it like that

S=\frac{1}{2}\sum_{k=1}^{+\infty}\frac{1}{2k(2k-1)}=\frac{1}{2}\sum_{k=1}^{+\infty} \left(\frac{1}{2k-1}-\frac{1}{2k}\right)=\frac{1}{2}\left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...\right]= \frac{1}{2}\ln 2

,where i made use of the second sum...

Daniel.

 

[saltydog]

The last is very easy,if u consider the Taylor series of \ln(1+x) around zero...(there's another elegant construction,too).

As for the first,write it like that

S=\frac{1}{2}\sum_{k=1}^{+\infty}\frac{1}{2k(2k-1)}=\frac{1}{2}\sum_{k=1}^{+\infty} \left(\frac{1}{2k-1}-\frac{1}{2k}\right)=\frac{1}{2}\left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...\right]= \frac{1}{2}\ln 2

,where i made use of the second sum...

Daniel.

Oh Jesus. I think you've already told me something like that before with another one. I'll spend some time with it. Thanks.