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monotonicity

Lisa91

New member
Jan 15, 2013
29
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.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
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.
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.
 

Lisa91

New member
Jan 15, 2013
29
I end up with [tex] e^{2x} \cdot e^{2} > x^{x} [/tex]. Is it ok? What shall I do with it?

Does the limit of this guy equal zero?
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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 $x = e^2 + 2$?
 

Lisa91

New member
Jan 15, 2013
29
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?
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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?
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.
 

Lisa91

New member
Jan 15, 2013
29
Do think wolfram alpha is wrong in this case?
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Do think wolfram alpha is wrong in this case?
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.
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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?
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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