Register to reply 
For which positive real numbers a does the series converge 
Share this thread: 
#1
Jan2612, 07:55 AM

P: 61

1. The problem statement, all variables and given/known data
For which positive real numbers a does the series Ʃo→∞ a^log(n) converge. Here logarithms are to the base e 2. Relevant equations Im afraid I'm not sure where to start, I'm not sure which topics would be applicable to this question. If someone could point me in the right direction or give me a clue, it would be much appreciated. tthanks 3. The attempt at a solution 


#2
Jan2612, 08:07 AM

P: 61

Is it possibly between 1 and 1? As log(n) will keep getting larger as n does. So if a<1 and a>1 then it will keep increasing in size so therefore it must be 1<=a=<1? If this is right how would I go about proving it?



#3
Jan2612, 08:08 AM

P: 61

thanks i'll have a look at that



#4
Jan2612, 08:09 AM

Emeritus
Sci Advisor
HW Helper
PF Gold
P: 7,800

For which positive real numbers a does the series converge
Write a as e^{log(a)}. Then using properties of logarithms you can show that this is a geometric series. BTW: The summation needs to be from 1 to ∞ . log(0) is undefined. 


#5
Jan2612, 08:29 AM

P: 61

so you mean
sum 1 to infinity e^(log(a))^(log(n)) I'm not sure what property i am trying to find. is there any need to change log(n) to the integral from 1 to n of 1/x? Or is there something to simplify log(a)^log(n). Otherwise i don't see the benefit of changing a to e^log(a). thanks 


#6
Jan2612, 08:51 AM

P: 61

is it
sum 1 to infinity of e^log(a^n)= sum 1 to infinity of a^n? or is that just rubbish? 


#7
Jan2612, 09:15 AM

Emeritus
Sci Advisor
HW Helper
PF Gold
P: 7,800

Perhaps I should have said properties of exponents and logarithms.
[itex]\displaystyle \left(b^k\right)^t=b^{kt}=\left(b^t\right)^k[/itex] 


#8
Jan2612, 09:15 AM

Sci Advisor
HW Helper
Thanks
P: 25,250




#9
Jan3112, 06:27 AM

P: 61

Thanks so i am left with
sum 1 to infinity n^(log (a)) correct? In regards to it converging im assuming a must be between 1 and 9 for this to converge as log1=0 and log9=0.954..., bigger than 9 then it wont converge but increase exponentially? 


#10
Jan3112, 09:16 AM

Sci Advisor
HW Helper
Thanks
P: 25,250




#11
Jan3112, 10:37 AM

Emeritus
Sci Advisor
HW Helper
PF Gold
P: 7,800

[itex]\displaystyle\sum_{n=1}^\infty\ n^{p}[/itex] converges for what values of p ?



#12
Feb112, 05:33 AM

P: 61

so ∑n=1,∞ n^p converges when 1<p<1 with p≠0, is that correct?



#13
Feb112, 05:45 AM

P: 61

so therefore a=9,8,7,6,5,4,3,2,2,3,4,5,6,7,8,9?



#14
Feb112, 08:54 AM

Sci Advisor
HW Helper
Thanks
P: 25,250




#15
Feb112, 09:02 AM

P: 61

is it nearly correct?
i dont think it will converge if bigger than 1 or 1 as it will increase exponentially, so it must be between them? 


#16
Feb112, 09:07 AM

Emeritus
Sci Advisor
HW Helper
PF Gold
P: 7,800

Perhaps it's more familiar to you in the following form. For what value of r, will the following converge? [itex]\displaystyle\sum_{n=1}^\infty\ \frac{1}{\ n^r}[/itex] 


#17
Feb112, 09:12 AM

P: 61

does it converge if p<1



#18
Feb112, 09:13 AM

P: 61

so it converges when r>1?



Register to reply 
Related Discussions  
Find all real numbers such that the series converges..  Calculus & Beyond Homework  19  
Real Numbers vs Extended Real Numbers  Linear & Abstract Algebra  5  
Show that R^x/<1> is isomorphic to the group of positive real numbers under multipli  Calculus & Beyond Homework  4  
Series converge/ diverges. determine sum of series  Mathematics Learning Materials  3  
Is there a way of proving that all positive numbers have a real square root?  Calculus & Beyond Homework  5 