Series

[daniele002]

Hey! I have no idea how to find the character of this series.
alpha is in R.

I've tried using the root test but it just equals 1, I've tried using asympotic relationships but n^5 doesn't converge to 0 so it was mistake. I have no idea at this point!
Thank you in advance!!

 

[Steven G]

What happens if alpha > 4/3??
For n > 1, 1+n^5 < n^5 + n^5 = 2n^5. Also ln(n) < n for n>0. Does this help?

 

[daniele002]

What happens if alpha > 4/3??
For n > 1, 1+n^5 < n^5 + n^5 = 2n^5. Also ln(n) < n for n>0. Does this help?

hey! thank you for replying
so would it diverge for alpha<4/3 and converge for alpha>4/3?

 

[Steven G]

hey! thank you for replying
so would it diverge for alpha<4/3 and converge for alpha>4/3?

I do not know. What do you think? Why would it converge for alpha > 4/3? More importantly why would you think that it converge if alpha <4/3. Why did you leave out the case when alpha = 4/3?

 

[daniele002]

I do not know. What do you think? Why would it converge for alpha > 4/3? More importantly why would you think that it converge if alpha <4/3. Why did you leave out the case when alpha = 4/3?

for alpha>4/3
(log(1+n^5)+n^3a+n^2)/(n^4+2)≤(2n^5+n^3a)/(n^4)

Now I'd write the numerator as n(2+n^(3a-5)) but I'm kind of stuck here.

 

[Steven G]

How did you get (log(1+n^5)+n^3a+n^2)/(n^4+2)≤(2n^5+n^3a)/(n^4)

 

[daniele002]

How did you get (log(1+n^5)+n^3a+n^2)/(n^4+2)≤(2n^5+n^3a)/(n^4)

if ln(n)<n and 1+n^5<2n^5 then ln(1+n^5)<2n^5 right?

 

[Steven G]

Compute the log of both sides of 1+n^5<2n^5. Why is that better then using ln(1+n^5)<2n^5?

 

[daniele002]

Compute the log of both sides of 1+n^5<2n^5. Why is that better then using ln(1+n^5)<2n^5?

I’ve tried another way and this is what I got. It makes sense to me because the log isn’t relevant for n approaching infinity, but I’m still not 100% sure.

 

[Steven G]

If alpha is 4/3 then the terms are (n^4 + n^2)/n^4 >1!

You did not show your work so I do not know if what you did was correct. I would not be very favorable if I were grading this.