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- Thread starter calcboi
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- Feb 21, 2013

- 739

Hello,

When it says lim x->1 it means x goes to 1 but not actually equal to 1. We aproximite. I would like someone to Approve that what I say is correct. I am sure but when it comes to help other i somehow get uncertain

Edit: If you want less Then 1 I think you Will lim x->1 (negative way)

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- Feb 13, 2012

- 1,704

The ratio test can be applied to the series with only positive or negative terms. Is Your series of this type?...

Kind regards

$\chi$ $\sigma$

- Jan 29, 2012

- 661

As has already been said, the ratio test is not convenient here. The series is $\displaystyle\sum_{n=1}^{\infty}\frac{\cos n\pi}{n^{2/3}}=\sum_{n=1}^{\infty}\frac{(-1)^n}{n^{2/3}}$, so the series of the absolute values is divergent (Riemann's series with $p=2/3\leq 1$). On the other hand by Leibniz criterion, the series is convergent. This implies that the given series is conditionally convergent.I am trying to determine convergence for the series n=1 to infinity for cos(n)*pi / (n^2/3)

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- Jan 17, 2013

- 1,667

It means we are choosing some \(\displaystyle |x-1|<\delta\) , which is a more general definition whether we are approaching from the right or the left. We can not say we are approximating the values of x since approximation has always a space of error.Hello,

When it says lim x->1 it means x goes to 1 but not actually equal to 1. We aproximite.