Prove the following sequence converges

Question

I am fairly confident that if $\alpha\in \mathbb{R}$ is such that $0<\alpha<1$, then the sequence $$(a_n):a_n=n\alpha^n$$ converges to $0$. I created a generalization of a method found in Prove $ne^{-n}$ converges to zero for $0<\alpha<1/2$ in which you argue $$n\alpha^n\leq \left(\frac{2}{(1/\alpha)}\right)^n$$ and use results about geometric series, but I am completely lost on the formal proof for $1/2<\alpha <1$. Maybe a hint to help me get started?

Alternatively, if such a hint/solution is overly complicated, I would greatly appreciate a simpler solution (if it exists) that can at least show the sequence is bounded.

Kavi Rama Murthy · Accepted Answer · 2020-07-10T06:20:01.077

1

$e^{nt} \geq \frac {n^{2}t^{2}} 2$ for $t \geq 0$. Put $t =-\ln \alpha$. You get $\alpha^{-n} \geq cn^{2}$ for some $c>0$. Can you finish?

edited Jul 10 '20 at 06:20

answered Jul 10 '20 at 05:57

Kavi Rama Murthy

311,013

If I am not mistaken, you are getting the fact that if $t>0$, then $e^{nt}\geq \frac{n^2t^2}{2}$ from the taylor series expansion of $e^x$? – northcity4 Jul 10 '20 at 06:02
@northcity4 Exactly. – Kavi Rama Murthy Jul 10 '20 at 06:03
Just saw your edit, that explains a lot. – northcity4 Jul 10 '20 at 06:04
@northcity4 There was a typo. I meant to take $t=- ln \alpha$ – Kavi Rama Murthy Jul 10 '20 at 06:07
Ok I think I got it. Given $\alpha$, we have shown that for all $n\geq 1$ that $n\alpha^n\leq 1/(nc)$. From here we can use something like a comparison test for sequences (such as the squeeze theorem with the $0$-sequence and the sequence $(1/nc)$). Is this correct? – northcity4 Jul 10 '20 at 06:11
@northcity4 Yes, you are right. – Kavi Rama Murthy Jul 10 '20 at 06:17
I should note for others seeing this post that the solution can be derived using the ratio test (only saw this later on). – northcity4 Jul 16 '20 at 08:50

Prove the following sequence converges

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