# Math Help - divergent series

1. ## divergent series

somebody know the value of the following divergent serie

sum[(-1)^m e^m,{1,Infinity}]

2. ## Re: divergent series

A series is called "divergent" if and only if it has no value!

3. ## Re: divergent series

it can be divergent but summable (C,K) ect..

4. ## Re: divergent series

Originally Posted by capea
it can be divergent but summable (C,K) ect..
You can only sum a finite number of terms of a divergent series.

5. ## Re: divergent series

I believe capea is talking about "Cauchy" or "mean" summability. capea, what do your "C" and "K" refer to?

6. ## Re: divergent series

Originally Posted by capea
somebody know the value of the following divergent serie

sum[(-1)^m e^m,{1,Infinity}]
A quick look suggests that this is neither Abel nor Cesaro summable

CB

7. ## Re: divergent series

Its an alternating series and by alternating series test it is divergent. Therefore it can not be calculated

8. ## Re: divergent series

for example the Sum((-1)^n Log(n),{n,1,Infinite}) is a divergent serie but is sumable and his value it is -1/2 Log(Pi/2),

9. ## Re: divergent series

Originally Posted by capea
for example the Sum((-1)^n Log(n),{n,1,Infinite}) is a divergent serie but is sumable and his value it is -1/2 Log(Pi/2),
Um, no it's not. The value partial sum value is here, and since it is divergent, it does not have a sum.

10. ## Re: divergent series

Originally Posted by capea
for example the Sum((-1)^n Log(n),{n,1,Infinite}) is a divergent serie but is sumable and his value it is -1/2 Log(Pi/2),
What capea wrote is 'almost true'... in the sense I try to explain now...

The so called 'Dirichlet eta function' is defined as...

$\eta (s)= - \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{s}} = (1-2^{1-s})\ \zeta(s)$ (1)

Writing the (1) as...

$\eta (s)= \frac{1}{2} + \frac{1}{2}\ \sum_{n=1}^{\infty} (-1)^{n-1}\ \{n^{-s} - (n+1)^{-s}\}$ (2)

... You can, with a little of patience, demonstrate that (2) converges for $\text{Re}\ s > -1$ and that is...

$\eta^{'} (0)= \frac{1}{2}\ \ln \frac{\pi}{2}$ (3)

Now if You compute 'directly' the derivative of (1) You obtain...

$- \eta^{'} (s) = \sum_{n=1}^{\infty} \frac{(-1)^{n}\ \ln n}{n^{s}}$ (4)

... so that setting in (4) $s=0$ and considering (3) You obtain...

$\sum_{n=1}^{\infty} (-1)^{n} \ln n = - \frac{1}{2}\ \ln \frac{\pi}{2}$ (5)

Kind regards

$\chi$ $\sigma$

11. ## Re: divergent series

for example

1^-s log(1)-2^-sLog(2).....==(2^(1-s)-1)Zeta´(s)-2^(1-s)Log(2 )Zeta(s)) so when s=0
you get the result