How would I show that: $$\sum_{n=1}^{\infty} \frac{y^2}{(1+y^2)^n}$$
converges for all $|y| \leq 1$ but that this does not converge uniformly?
I wrote out the series which is:
$$\frac{y^2}{(1+y^2)} + \frac{y^2}{(1+y^2)^2} + \frac{y^2}{(1+y^2)^3} + ... $$ $$=\frac{y^2}{(1+y^2)}*(1 + \frac{1}{(1+y^2)} + \frac{1}{(1+y^2)^2} + ...)$$
$$=\frac{y^2}{(1+y^2)}*(\frac{1}{1 - (\frac{1}{(1+y^2)})})$$ $$ = 1$$
So this series converges to $1$ by the geometric series, now how would I show that this doesn't converges uniformly? I know the definition of uniform convergence which pretty much says
$\forall \epsilon >0,$ there is an $N > 0$ so $\forall y$ and $n \geq N$ we have $|f_n(y) - f(y)| < \epsilon$
But from this, I can't really think of an $\epsilon$... So how would I really approach this problem?
Note that the sequence of function $$f_n(y) = \sum_{k=1}^n \dfrac{y^2}{(1+y^2)^n} = 1 - \dfrac1{(1+y^2)^n}$$ Clearly, $f_n(y) \to 1$. However, not uniformly, since given any $\epsilon > 0$ and any $n \in \mathbb{N}$, we can always find $y \in [-1,1]$ such that $\dfrac1{(1+y^2)^n} > \epsilon$. To be specific, we can choose any $y < \sqrt{\dfrac1{\epsilon^{1/n}}-1}$, if $\epsilon < 1$ and to be any number $y<1$ for $\epsilon \geq 1$.