Convergence of the series $w_n = (-1)^{n}x^{n}(1-x)$ on $(0,1)$, then $\sum w_n$ is uniformly convergent.
Idea : I wanted to use the M-test for convergence of series,so I found out the value of x for which $w_n$ is maximum on (0,1) but the value comes as an expression in n where the sign of the such an x also depends on n, so I am stuck here.
Is the direction of my approach correct?
You don't need Weierstraß' M-test here: $\lvert w_n(x)\rvert$ has a maximum $M_n$ for $x=\dfrac n{n+1}$ and this maximum is $$M_n=\Bigl(\frac{n}{n+1}\Bigr)^n\Bigl(1-\frac n{n+1}\Bigr)\le\frac1{n+1}.$$ Now Leibniz's test for alternating series says the remainder at rank $n$ satisfies $$\lvert R_n(x)\rvert\le\lvert w_{n+1}(x)\rvert\le M_{n+1}=\frac1{n+2},$$ which proves the convergence is uniform on $(0,1)$.