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monotonicity
- Thread starter Lisa91
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- #2
- Feb 7, 2012
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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.
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- #3
- Feb 15, 2012
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what you are looking for is $x$ such that:
$2x + 2 < x\log(x)$ (you want $f$ to be DECREASING).
note that if $x>e$ the RHS is increasing faster than the LHS, so if we find some $x>e$ with the above inequality true, the LHS will never "catch up" for larger $x$.
one $x$ we might try is:
$x = e^2$, unfortunately:
$2e^2 + 2 > 2e^2$.
how about $x = e^2 + 2$?
$2x + 2 < x\log(x)$ (you want $f$ to be DECREASING).
note that if $x>e$ the RHS is increasing faster than the LHS, so if we find some $x>e$ with the above inequality true, the LHS will never "catch up" for larger $x$.
one $x$ we might try is:
$x = e^2$, unfortunately:
$2e^2 + 2 > 2e^2$.
how about $x = e^2 + 2$?
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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?
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?
- Feb 5, 2012
- 1,621
Hi Lisa91,
Refer Opalg's post #2 and you can show that \(\frac{\ln(n)}{\sqrt{n+1}}\) is a decreasing sequence of positive numbers. Then by the Leibniz test the series \(\sum (-1)^n \frac{\ln(n)}{\sqrt{n+1}}\) converges.
Kind Regards,
Sudharaka.
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- #7
- Feb 5, 2012
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Yes, wolfram is not always correct.
Can you link me to the answer that wolfram gave? >>Here<< is what I got using wolfram. Note that it doesn't say anything about convergence/divergence.
- Feb 15, 2012
- 1,967
how so? you want $2x+2$ to be dominated by $x\log(x)$, for this means that:
$\displaystyle f'(x) = \frac{2x+2 - x\log(x)}{(2x)(x+1)(\sqrt{x+1})} < 0$
that is, f is decreasing.
in fact: $e^2 + 2 \sim 9.389$ so the sequence:
$\displaystyle \frac{2\log(n)}{\sqrt{n+1}}$
is definitely decreasing for $n \geq 10$
my input for the series you seem to actually want on wolframalpha is here:
sum from n=1 to infinity (-1)^n(2log(n))/((n+1)^(1/2)) - Wolfram|Alpha
it merely says the results are inconclusive, but if you look at the partial sums for 600 terms, you can see it's very SLOWLY starting to converge.
by the way the function:
$\displaystyle f(x) = \frac{2\log(x)}{\sqrt{x+1}}$
is quite "flat", so the convergence of your original sequence (and it does converge, to 0, in fact) is very slow.
my guess is that it (the alternating series) converges to around 0.127 or so....
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