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## richyw Group Title limits. How do I show that $\lim_{k\to\infty} \frac{(-1)^k}{k}=0$ one year ago one year ago

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1. shubhamsrg

Do you know LH rule ?? Where you can differentiate the numerator and denominator.

2. richyw

sorry I keep messing up the latex and openstudy is so slow.

3. richyw

and yeah I know lh rule could be applied since $$k\in\mathbb{R}$$. But how does that help?

4. shubhamsrg

Okay since it is (-1) , LH rule won't be applied.

5. shubhamsrg

hmm, so, think like this : (-1)^n , where n is any natural number, will always be equal to +1 or -1 right ?

6. richyw

agreed. and since $$k\to\infty$$ I can see that it's 0. just not sure how to show that in any sort of rigorous way.

7. richyw

I need to show it though to show that a series is convergent.

8. shubhamsrg

hmm, Am not too comfortable with that. @phi

9. phi

I thought the definition of a limit L is that for any $$\epsilon >0$$, there exists an integer N>0 such that $$| L - x_n| < \epsilon$$ for all n>N the alternating sign does not affect this definition, because you are only looking at the distance away from L. (0 in this case)