Does $\frac {1^\alpha +2 ^\alpha+...+n^\alpha} {n^{\alpha+1}}$ diverge to $\infty$?

Question

Let $\alpha \in ( - \infty,-1)$. Does $\frac {1^\alpha +2 ^\alpha+...+n^\alpha} {n^{\alpha+1}}$ diverge to $\infty$ as $n \to \infty$?

(I am not 100% sure that the above is true, but I am pretty sure because I am trying to use it as a lemma to show that a certain Lebesgue integral is $\infty$).

My thought was to try bounding $\frac {1^\alpha +2 ^\alpha+...+n^\alpha} {n^{\alpha+1}}$ below by something which is already known to diverge to $\infty$.

I see $\frac {1^\alpha +2 ^\alpha+...+n^\alpha} {n^{\alpha+1}}\geq\frac {n^\alpha +n ^\alpha+...+n^\alpha} {n^{\alpha+1}}=1$, but this doesn't really help because the bound $1$ is too small to be useful.

The formula $1^3+2^3+...+n^3=(1+2+...+n)^2$ comes to mind, but I'm not sure how it would help since $\alpha \neq 3$. (Maybe the formula can be manipulated somehow to apply in our situtation? Or maybe there is a more general formula I can use?)

Any help would be appreciated.

I suggest the lower bound $\frac 1 {n^{\alpha+1}}.$ For the given range of alpha, this grows with n. — Artimis Fowl, Dec 05 '19 at 11:34
This is similar to this question and all the ones linked to it, but the range of the parameter is not always clear. — Arnaud D., Dec 05 '19 at 11:37
This answer would work in your case, but if that's how you got the question in the first place I suppose it won't help. — Arnaud D., Dec 05 '19 at 11:45

score 1 · Answer 1 · answered Dec 05 '19 at 11:37

1

The numerator tends to $\zeta(-\alpha)>0$ which is finite since $-\alpha>1 $.

The denominator tends to $0^+$.

Therefore your lemma is true.

answered Dec 05 '19 at 11:37

Arnaud Mortier

27,276

Does $\frac {1^\alpha +2 ^\alpha+...+n^\alpha} {n^{\alpha+1}}$ diverge to $\infty$?

1 Answers1