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Let $f_n(x) = n^\alpha x^n(1-x)$ for any fixed $\alpha \in \mathbb{R}$ and $x \in [0,1]$. Prove $f_n$ converges to $0$ pointwise (via the definition).

I considered that
$|n^\alpha x^n(1-x) - 0| = n^\alpha x^n (1-x) \leq n^\alpha x^n$.
But then I'm stuck here, because it seems that I can't isolate the $n$ term, to get a relationship between $n$ and $\epsilon, x$.

Natash1
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  • I think you can safely assume that $x<1$. The $x=1$ case really should not cause difficulty. – Angina Seng Jun 03 '17 at 13:01
  • Oh right....... that means I can simplify to $\leq n^\alpha$ ... (which also implies that this function is uniform converging?!) – Natash1 Jun 03 '17 at 13:06
  • Simplifying to $n^{\alpha}$ wouldn't help much, because $\alpha$ can be any real number, right? – SvanN Jun 03 '17 at 13:07
  • Why couldn't I just let $N:= \lfloor \epsilon ^{\frac{1}{\alpha}} \rfloor + 1$ because if $n^\alpha < \epsilon \iff n < \epsilon ^{\frac{1}{\alpha}}$? for non-zero $\alpha$ – Natash1 Jun 03 '17 at 13:09
  • You know, when I see a question that asks to prove that a sequence is ''pointwise convergent" I generally suspect that it is not uniformly convergent. – Angina Seng Jun 03 '17 at 13:19
  • Hmm I see. I'm still unsure how to continue from my step in the pointwise proof – Natash1 Jun 03 '17 at 13:22
  • How would you approach this problem: prove that $\sum_{n=1}^\infty n^\alpha x^n$ is convergent when $|x|<1$? – Angina Seng Jun 03 '17 at 13:24
  • @S.vanNigtevecht Nevermind, I see your point, now. – Natash1 Jun 03 '17 at 13:24
  • Would we use the ratio test? or maybe integral test since we want an inequality somewhere? – Natash1 Jun 03 '17 at 13:25

1 Answers1

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The cases $x=0$ and $x=1$ are trivial, so we can limit ourselves to $x \in (0,1)$. The factor $1-x$ is simply a constant and is irrelevant to convergence. As such, all we need is $n^{\alpha} x^n$ to go to zero as $n$ goes to infinity. A 'trick' for this could be to use the ratio test on $\sum n^\alpha x^n$:

$$\dfrac{(n+1)^\alpha x^{n+1}}{n^\alpha x^n} = \left(\dfrac{n+1}{n}\right)^\alpha \cdot x.$$

Note how $\lim_{n\to\infty} \frac{n+1}{n} = 1$, so because $|x| < 1$, we have: $$ \limsup_{n\to\infty} \left | \dfrac{(n+1)^\alpha x^{n+1}}{n^\alpha x^n} \right | < 1,$$ so the series $\sum n^\alpha x^n$ converges. This means its individual terms go to zero, so we finally have $n^\alpha x^n \to 0$ as $n \to \infty$.

SvanN
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