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For a real number $x$ such that $|x|<1$, show that the sequence $(x^n\sqrt{n})_{n \in \mathbb{N}}$ is convergent, with limit 0.

Here's what I have so far: Let $\epsilon > 0$. We want to find $N$ such that for all $n > N, |x^n\sqrt{n}| < \epsilon$. Now we have $n\log(x) + \frac{1}{2}\log(n) < n + n = 2n < \log (\epsilon)$. So pick $N$ with $N<\frac{\log(\epsilon)}{2}$.

My question is, is using logarithms a valid way to approach a question on convergence, and is there a better alternative method to tackle this question?

EDIT: Attempt 2: If $x=0$, then $x^{n}\sqrt{n}=0$ for every $n \in \mathbb{N}$ and therefore $$\lim_{n\to\infty}x^{n}\sqrt{n}=0$$ So we can assume $x \neq 0$. Then, note that $h = \frac{1}{|x|} - 1 > 0$. From Bernoulli's inequality we have: $\frac{1}{|x|^{n}\sqrt{n}} = \frac{(1+h)^{n}}{\sqrt{n}} \geq \frac{1+nh}{\sqrt{n}} > \sqrt{n}h.$ Therefore, we have $0 \leq |x|^{n}\sqrt{n} \leq \frac{1}{\sqrt{n}h}.$ Now, given any $\epsilon > 0$, we take $N \geq (\frac{1}{\epsilon h})^2$. Now for $n>N, |x^n\sqrt{n} - 0| = |x^n\sqrt{n}| \leq \frac{1}{\sqrt{n}h} < \frac{1}{\sqrt{N}h} \leq \epsilon$. Hence $$\lim_{n\to\infty}x^{n}\sqrt{n}=0$$

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    Even if $N<\log(\epsilon)/2$, there will be some $n>N$ for which $2n>\log(\epsilon)$, so your choice of $N$ doesn't work. Also, for small $\epsilon$, we'll have $\log(\epsilon)<0$, so you won't be able to even pick your initial $N$. – 2017-01-29
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    Yeah you're right. I'm a bit stuck then, how should I approach this question? – 2017-01-29
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    Suggestion: Show that $(\log n)/n\to0$ as $n\to\infty$. Then note that $\log |x|<0$ because $|x|<1$; then look at $n\log |x|+{1\over2}\log n$ and see it go to $-\infty$ as $n\to\infty$. – 2017-01-29

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Raise to square, and take limit for $n \to \infty$

Then use L'Hopital rule

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    Formally we haven't covered L'Hopital's rule. Is there another way to approach this from the definition of convergence? – 2017-01-29
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Note that since $x\mapsto x^2$ is continuous $x^n\sqrt n\to 0$ if and only if $nx^{2n}\to 0$. We have

$$\sum_{n=0}^\infty nx^{2n} = \frac 12\sum_{n=0}^\infty 2nx^{2n} = \frac 12x\left(\sum_{n=0}^\infty x^{2n}\right)' =\frac 12 x\left(\frac{1}{1-x^2}\right)'=\frac{x^2}{(1-x^2)^2},\quad |x|<1$$

We used convergence of geometric series here. Since $\sum nx^{2n}$ converges, $nx^{2n}\to 0$ by well known necessary criterion for series convergence.

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    Taking derivatives of power series is too advanced to justify to a lot of beginning analysis students. – 2017-01-29
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If there exists $c\in(0,1)$ such that $\left|\dfrac{u_{n+1}}{u_n}\right| \le c$ for all $n$, then $u_n$ converges to $0$ (because $|u_n| \le |u_1|c^{n-1}$, and $c^{n-1}$ converges to $0$).

More generally, if there exist $c \in(0,1)$ and $N\in\mathbb N$ such that $\left|\dfrac{u_{n+1}}{u_n}\right| \le c$ for all $n \ge N$, then $u_n$ converges to $0$ (because $|u_n| \le |u_N|c^{n-N}$ if $n \ge N$, and $c^{n-N}$ converges to $0$).

So, given $x$ such that $|x|<1$, you want to find $c$ and $N$ such that $\left|\dfrac{u_{n+1}}{u_n}\right| = |x|\sqrt{\dfrac{n+1}{n}}$ is less than $c$ for all $n\ge N$. Can you do this?

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The attempt with the logarithm is interesting, albeit wrong.

Note that $|x|<1$, so $\log|x|<0$. Moreover also $\varepsilon$ might be less than $1$ (and you have to prove the assert for every $\varepsilon>0$).

The inequality $$ |x^n\sqrt{n}|<\varepsilon $$ becomes $$ n\log |x|+\frac{1}{2}\log n<\log\varepsilon $$ Since $\frac{1}{2}\log n\frac{\log\varepsilon}{\frac{1}{2}+\log|x|} $$ when $\log|x|<-1/2$, but no information when $-1/2\le\log|x|<0$. A stricter inequality than $\log n<2n$ should be used.

Using Bernoulli's inequality seems more promising. The inequality can be rewritten $$ \frac{1}{\varepsilon}<\frac{1}{|x|^n\sqrt{n}} $$ (we can assume $x\ne0$, because for $x=0$ there's nothing to prove). Set $1/|x|=1+t$, with $t>0$. Then $$ \frac{1}{|x|^n\sqrt{n}}=\frac{(1+t)^n}{\sqrt{n}}\ge\frac{1+nt}{\sqrt{n}} $$ If the inequality $$ \frac{1+nt}{\sqrt{n}}>\frac{1}{\varepsilon} $$ is satisfied for $n>N$ ($N$ depending on $\varepsilon$), we are done. Can you do it?