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Let $f(x) = \frac{2\ln x}{\sqrt{x+1}}$. Differentiate it, and show that $f'(x)<0$ whenever $x>9$. This will show that the sequence $\frac{2\ln(n)}{\sqrt{n+1}}$ decreases from $n=9$ onwards. Notice that this sequence does not decrease to start with, in fact it increases from $n=1$ up to $n=8$. But that does not matter when investigating convergence, because convergence is unaffected by ignoring the first few terms.Could anyone help me out with the monoticity of this sequence please?
[tex] \frac{2\ln(n)}{\sqrt{n+1}} [/tex]. It should decline. I am investigating the convergence of one series and I need it to do the Leibniz test.
Hi Lisa91,When [tex] x = e^{2} +2 [/tex] we get [tex] 2x+2< x \ln(x) [/tex]. So, this is not what I want.
Ok, I thought I could solve it on my own but it seems to be much more complicated. I have a series [tex] \sum (-1)^n \frac{2\ln(n)}{\sqrt{n+1}} [/tex]. I thought I could use the Leibniz test but even wolfram alpha shows that the series doesn't converge. Could you give any hints how to prove the divergence of this series?
Yes, wolfram is not always correct.Do think wolfram alpha is wrong in this case?
how so? you want $2x+2$ to be dominated by $x\log(x)$, for this means that:When [tex] x = e^{2} +2 [/tex] we get [tex] 2x+2< x \ln(x) [/tex]. So, this is not what I want.
Ok, I thought I could solve it on my own but it seems to be much more complicated. I have a series [tex] \sum (-1)^n \frac{2\ln(n)}{\sqrt{n+1}} [/tex]. I thought I could use the Leibniz test but even wolfram alpha shows that the series doesn't converge. Could you give any hints how to prove the divergence of this series?