Please check my attempt and result in solving the following question:

Let $f_n(x)= x^n(1-x)^2$, $x\in [0,1]$.

Does $\sum f_n$ converge? If so, does it converge normally? uniformly?

My attempt:

Pointwise convergence:

Using D'Alembert (ratio) test: Let $x^n(1-x)^2= a_n$.

$|\frac{a_{n+1}}{a_n}|= |x|$ $\leq 1$. If we try to substitute for $x=1$ in the series, we get $\sum 0$ which's a convergent series. So by ratio test we have pointwise convergence.

Normal convergence:

After calculating the derivative of $f_n(x)$, I got that the function attains its maximum at $x= \frac{2n}{2n+4}$. And the function is obviously bounded.

$$\|f_n\|_\infty = \frac{1}{e^2}\left(\frac{4}{2n+4}\right)^2.$$

$$\sum_{k=0}^\infty \|f_k\|_\infty = \sum_{k=0}^\infty \frac{1}{e^2} \left(\frac{4}{2n+4}\right)^2 = \sum_{k=0}^\infty \frac{16}{4k^2e^2 + 16e^2 + 16ke^2} < \sum_{k=0}^\infty \frac{1}{k^2}.$$

By comparison test, we get that $\sum_{k=0}^\infty \|f_k\|_\infty$ converges. Therefore, the series normally converges and so uniformly.

I don't know if I'm working well, but I tried another way I may use it if I was not asked about normal convergence (as I have an exam after 3 days and I wanna be strong at such problems).

My other way:

After proving the pointwise convergence, I thought of proving that $\|R_n\|_\infty \longrightarrow 0$:

$|R_n(x)|= \sum_{k=n+1}^\infty x^k(1-x)^2 \leq \sum_{k=n+1}^\infty x^k$, which converges for $x\in [0,1)$, and for $x=1$, we talked about it above. Hence, $|R_n(x)| \leq c$ for some positive $c$ (i.e it's bounded).

Proving that $R_n$ is a decreasing sequence:

$R_{n+1}(x) - R_n(x)= -f_{n+1}$, and we have $f_n \geq 0$, $R_{n+1}(x) - R_n(x) \leq 0$ and so it's decreasing. Same for $|R_n(x)|$. We have now $|R_n(x)|$ is bounded and $\leq c$ $\forall x$, particularly we have $\|R_n\|_\infty$ $\longrightarrow 0$. Therefore we have uniform convergence.

Am I right in both ways? One of them? Any advices are appreciated. By the way, I didn't took Weirestrass M-test as I see that many problems can be solved using it, but unfortunately I can't. So please give me advices for these 2 ways as I didn't took any other way.

  • 1
    $\begingroup$ Rather than applying the ratio test, you could say that the series is geometric, i.e. it has a common ratio. The common ratio is $x$, and you can say exactly what the sum of the series is, since there is a simple formula for the sum of a geometric series. The case $x=1$ must be treated separately and you've done that already: In that case every term of the series is $0.$ $$ \sum_{n=0}^\infty x^n (1-x)^2 = \sum_{n=0}^\infty ar^n = \frac a {1-r} = \frac{(1-x)^2}{1-x} = 1-x. $$ $\endgroup$ – Michael Hardy May 20 '17 at 15:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.