1
$\begingroup$

How can I use Weierstrass test to determine a set $I$ on which the series $\displaystyle \sum_{n=1}^{\infty} nx^n(1-x)^n$ converges uniformly?

Weierstrass test: Let $f_n : I \to \mathbb{R}$ be a sequence of functions with $|f_n(x)| \le M_n$ for all $x \in I$ and $k=1,2, \dots$. If $\sum_n M_n$ converges then $\sum_n f_n$ converges uniformly on $I$.

$\endgroup$
  • $\begingroup$ Is your series $\sum n x^n(1-x)^n$? $\endgroup$ – David Mitra Jan 5 '12 at 15:27
  • $\begingroup$ Yes that is the series $\endgroup$ – neemy Jan 5 '12 at 15:36
  • 1
    $\begingroup$ Defining $g(x)=\sum_n nx^n$, we can see that $f(x)=g(x(1-x))$. $\endgroup$ – Thomas Andrews Jan 5 '12 at 16:22
4
$\begingroup$

The Weierstrass test is quite intuitive: it tells you that if you can bound each function in the series by a constant, and that this new series of constants converges, then so does your sum. (And what's more, it converges uniformly.)

First note that if $\sum_n M_n$ is to converge then the sequence $(M_n)$ must be bounded. And since each $M_n$ is a bound for $f_n$, this means that the sequence $(f_n)$ must be uniformly bounded on $I$, the domain of the functions in the sequence.

So there are two steps that you need to follow. First you need to find a subset $I \subseteq \mathbb{R}$ on which you can uniformly bound the $f_n$. Then you need to find whether the bounds you can attain satisfy the convergence condition of the Weierstrass test.

Now, for any compact $I \subseteq \mathbb{R}$, your $f_n$ will be bounded in $x \in I$ because they are continuous. So you need to look instead at $n$: for which subset $I \subseteq \mathbb{R}$ is the sequence $(f_n(x))$ bounded for each fixed $x \in I$?

Once you have worked this out, you can test to see if this subset gives way to the constants $M_n$ for which $|f_n(x)| \le M_n$ for each $x \in I$ and for which $\sum_n M_n$ converges.

$\endgroup$
  • 2
    $\begingroup$ I'll just add, if I may, that the series $\sum\limits_{n=1}^\infty n r^n $ converges if and only if $|r|<1$ $\endgroup$ – David Mitra Jan 5 '12 at 16:10
1
$\begingroup$

So, your series being $\displaystyle \sum_{n=1}^{\infty} nx^n(1-x)^n$, I take it that you have to prove the uniform convergence on the interval $[0,1]$ (tell me if I am wrong). In this interval, the function $f(x)=x(1-x)$ is positive and its maximum $1/4$ is taken at $1/2$ (where $f'(x)=0$). Hence $$ nx^n(1-x)^n\leq \frac{n}{4^n}=M_n $$ Now, the sum $$ \sum_{n}\frac{n}{4^n}\leq \sum_{n}\frac{1}{2^n}<+\infty $$ which proves the convergence.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.