# Leibniz's rule application

I need to show whether the following serie converges or diverges:

$$\sum_{n=0}^{\infty} (-1)^n \frac{\sqrt{n}}{n+100}$$

I need to use ONLY Leibniz's rule. I started by writing:

$$\sum_{n=0}^{\infty} (-1)^{n-1}(- \frac{\sqrt{n}}{n+100})$$

Now I wanted to show that $- \dfrac{\sqrt{n}}{n+100}$ decreases but I am having trouble showing that.

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Well, since the sequence $\,\displaystyle{\frac{\sqrt n}{n+100}}\,$ is eventually monotone converging to zero (why?), Leibnitz theorem/test applies here and thus your series converges.

Oh, and it really doesn't matter, for convergence purposes, whether you being with a positive or a negative summand in the series.

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Can you detail a little more the monotony illustration please ? –  Carpediem Nov 29 '12 at 13:43
It is NOT converging monoton as you can see in my answer. The sequence is increasing from n=0 to 100 and decreasing monoton for $n>100$. –  user127.0.0.1 Nov 29 '12 at 13:47
Yeah, well: you're right, @Fant, but the text applies for eventually monotone sequences. Point taken though, thanks. –  DonAntonio Nov 29 '12 at 13:53
@Fant: You want just a starting point in this test. –  Babak S. Nov 29 '12 at 14:22
@BabakSorouh: I know. Anyhow the sequence is not monoton for all n. That's why I just gave a comment (so DonAntonio might improve his answer) and didn't downvote the answer or sth like that. –  user127.0.0.1 Nov 29 '12 at 14:26

You have $$\frac{d}{dn}\left(\frac{\sqrt{n}}{n+100}\right) = \frac{100-n}{2\sqrt{n}(n+100)^2}$$ which is $\lt 0 \ \ \forall n \gt 100$.

That's why $\frac{\sqrt{n}}{n+100}$ decreases monotonically for $n>100$.

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I hope, you accept my edit just for the derivation. It needs parentheses. –  Babak S. Nov 29 '12 at 14:32
Like your attempt. –  Babak S. Dec 1 '12 at 14:03