so I have a problem with 4 convergence questions. I have done the ones where I have to use other methods such as the ratio test and they were OK, but I cannot understand this one in particular.
The problem says: "Let $f_n(x)=x^n$ on $I=[0,1)$, does $\sum\limits_{n=1}^\infty f_n(x)$?"
And the answer says:
The geometric series converges pointwise to $s(x)=\frac{x}{1-x}$.
The sequence of partial sums is $s_n(x)=\frac{x(1-x^{n+1})}{1-x}$ so that $$s_n(x)-s(x)=\frac{x(1-x^{n+1})}{1-x}-\frac{x}{1-x}=-\frac{x^{n+2}}{1-x}.$$
And here is where I get it wrong:
If we take $x=1-1/n$ we get $$|s_n(1/n)-f(1/n)|=n(1-1/n)^2(1-1/n)^n$$ which is unbounded since $(1-1/n)^2(1-1/n)^n \to e^{-1}\neq 0$. Therefore, $\sup\limits_{x\in I}|s_n(x)-f(x)|$ does not converge to 0 as $n\to\infty$, so we do not have uniform convergence of the series.
Comments: So I know $f(x)=\lim_{n\to\infty}f_n(x)=0$, so when I let x=1-1/n I get $$|s_n(1-1/n)-f(1-1/n)|=|n(1-1/n)(1-(1-1/n)^{n+1})|=|n(1-1/n-(1-1/n)^2(1-1/n)^n)|$$
And this is very different from the actual result...
I would really appreciate it if you someone could clarify me how this last bit is done. Thanks!