Does it converges?

sinos

New member
Joined
Jul 15, 2019
Messages
5
Hello,
I have two question:
Firstly,
I got this problem: 12995

I should determine weather the series converges.
I thought about using leibinz test but I don't know how to prove that 12996
is monotonlly decreasing.

The second qeustion I would like to ask is, if I don't know about a series if its positive, can I use d'almbert ratio test in this way:
מבחן ד',למבר



and if for example for the none positive series I show in this way that the series diverges is this a legit proof?
Thank you!
 

Attachments

pka

Elite Member
Joined
Jan 29, 2005
Messages
8,445
I have two question: Firstly,
I got this problem: View attachment 12995

I should determine weather the series converges.
I thought about using leibinz test but I don't know how to prove that View attachment 12996
is monotonlly decreasing.
The second qeustion I would like to ask is, if I don't know about a series if its positive, can I use d'almbert ratio test in this way:
מבחן ד',למבר',למבר
Consider the function \(\displaystyle f(x)=\frac{\log^2(x)}{x^{1/3}}\) then \(\displaystyle f'(x)=-\frac{(\log(x)-6)\log(x)}{3x^{4/3}}\) SEE HERE
Can you see that the derivative is negative if \(\displaystyle x>e^6~?\)
Thus apply the alternating series test. Post your results. So we can proceed.

 

sinos

New member
Joined
Jul 15, 2019
Messages
5
Consider the function \(\displaystyle f(x)=\frac{\log^2(x)}{x^{1/3}}\) then \(\displaystyle f'(x)=-\frac{(\log(x)-6)\log(x)}{3x^{4/3}}\) SEE HERE
Can you see that the derivative is negative if \(\displaystyle x>e^6~?\)
Thus apply the alternating series test. Post your results. So we can proceed.

Hi,
thank you for your respond.
Why did you use log instead of lan? (I know lan is a private case of log but still I don't understand)

thank you
 

pka

Elite Member
Joined
Jan 29, 2005
Messages
8,445
Hi,
thank you for your respond.
Why did you use log instead of lan? (I know lan is a private case of log but still I don't understand)

thank you
It is current practice to use \(\displaystyle \log(n)\) for the so-called natural logarithm.
 

lookagain

Senior Member
Joined
Aug 22, 2010
Messages
2,447
Hi,
thank you for your respond.
Why did you use log instead of lan? (I know lan is a private case of log but still I don't understand)

thank you
If you got your problem for a course/from an instructor, then you answer it back with the same "ln" notation.

And, no, pka is not correct about it being (exclusive) current practice. His statement comes from his bias.
Your own question is testament to that. It is in mixed usage today.
 

sinos

New member
Joined
Jul 15, 2019
Messages
5
okay.
Thank you so much!
 
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