[itex](-1)^n[/itex] and [itex] \quad i^n[/itex] do not tend to zero, so the associate series is not convergent.hola said:Hey. This has been bugging me for a long time:
why does summation from n=1 to infinity of (-1)^n or i^n or 1/n or -1/n not converge, because summation from n=1 to infinity of 1/n^2 conveges. Don't the terms in (-1)^n or i^n or 1/n or -1/n(-1)^n tend to 0?
dextercioby said:Well,the trick is with the series
[tex] \sum_{k=1}^{+\infty}\frac{(-1)^{k+1}}{k} [/tex].
IIRC it converges to
[tex]\ln 2 [/tex].
Daniel.
dextercioby said:You can start from this series
[tex] \frac{1}{1+x}=1-x+x^{2}-x^{3}+x^{4}-... [/tex]
and integrate it term by term and you'll have to put x=1 in the new series to find the formula which I've hinted.
Daniel.
Some summations do not converge because the terms in the series do not approach a finite limit as the number of terms increases. This can happen when the terms in the series alternate in sign or increase too quickly.
There are a few methods for determining the convergence of a summation, including the comparison test, the ratio test, and the integral test. These methods involve comparing the given series to other known series that either converge or diverge.
No, a divergent summation cannot have a finite sum. If a summation is divergent, it means that the terms in the series do not approach a finite limit, so the sum of the series is also infinite.
A convergent summation has a finite sum, meaning that the terms in the series approach a finite limit as the number of terms increases. A divergent summation, on the other hand, does not have a finite sum as the terms in the series do not approach a limit.
Yes, a summation with an infinite number of terms can still converge if the terms in the series decrease in value quickly enough. This is known as a convergent infinite series. An example of this is the geometric series, where the terms decrease by a constant ratio.