# How to use Weierstrass test?

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 $I$.

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

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.

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

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.