Why does $\sum_{n=1}^\infty (-1)^nx^n(1-x) $ not have a convergent majorant series?

Question

I am working on this problem:

Find the sum of $f(x) = \sum_{n=1}^\infty (-1)^nx^n(1-x), x \in [0,1] $.

Why does $\sum_{n=1}^\infty (-1)^nx^n(1-x), x \in [0,1] $ not have a convergent majorant series?

I found the sum to be

$$\sum_{n=1}^\infty (-1)^nx^n(1-x) = \sum_{n=1}^\infty (-1)^nx^n - \sum_{n=1}^\infty (-1)^nx^{n+1} $$

$$\sum_{n=1}^\infty (-x)^n - x\sum_{n=1}^\infty (-x)^n $$

$$f(x) = \bigg(\frac{1}{1+x}-1 \bigg) -x \bigg(\frac{1}{1+x}-1 \bigg) = \frac{x(x-1)}{x+1} $$

But I can't figure out how to solve the second part of the question. Why does the sum not have a convergent majorant series?

Answer 1

If $|(-1)^{n}x^{n}(1-x)| \leq a_n$ for all $x$ and $n$ then, putting $x=\frac n {n+1}$, we get $(1+\frac 1 n)^{-n} \frac 1 {n+1} \leq a_n$. Use the fact that $(1+\frac 1 n)^{-n} \to \frac 1 e$ to see that $\sum a_n=\infty$.

How did I think of $x=\frac n {n+1}$? This is just the point where left side is maximized.