Can I say the function is decreasing since when p is negative the denominator is growing faster than the numerator so it is decreasing?
Ok so I'm working on this problem but have problem understanding it. I got the right answer but I made an assumption to get there. I need help proving that assumption.
Here is the problem
Find the values of p for which the series in convergent
Ok so I used the integral test because f(x) = is continous, positive, and decreasing(this is where I made the assumption that it was decreasing).
Thus the limit exist when so the series converges when p< -1.
Is that right? Also how do I show that the function is decreasing(the assumption I need to prove). I took the 1st derivative and equalled it to less than 0 to show me where it was decreasing but ran into problems.
This is what I get stuck on with the derivative
Thanks
Hello,
Yes, it is continuous.
Actually, my reasoning is as follows :
The series {a_n} converges ---> lim a_n = 0
By taking the contrapositive of this assumption, it's evident to say that lim a_n has to be 0.
Assuming p<0 (otherwise, one can see it is not possible), let q=-p for more convenience :
---> for it to tend to 0, q has to be > 1/2, which is p<-1/2. Necessary condition, not sufficient.
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If you want to show it's decreasing :
x(1+x²)^p
Its derivative is
because
We want to find p such that it is
if ---> (2p+1)x²+1 is always positive. Which is not possible.
if , since x²>1, (2p+1) has to be < -1, that is to say p<-1
etc.