to show a function is continuous Consider $f(x)=\sum_{n=1}^\infty n^2x^n ; x\in [-\dfrac{1}{2},\dfrac{1}{2} ]$ To show if $f$ is continuous therein. Now what I find is $f$ is pointwise convergent for each $x\in [-\dfrac{1}{2},\dfrac{1}{2} ]$ Now if we can show that the series is uniformly convergent then I can conclude that $f$ is continuous But cant find the sum of the given series.Any help on how to proceed

A power series converges uniformly on any compact interval within the radius of convergence. Use the ratio test to show that the radius of convergence is $1$ (i.e the series converges if $|x| < 1$).

Alternatively, apply the Weierstrass M-test noting that in $[-1/2,1/2]$ we have $|n^2x^n| \leqslant n^22^{-n}$. Show that the series of dominating terms converges using the ratio test.

With $a_n = n^22^{-n}$, we have

$$\lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right|= \lim_{n \to \infty}\frac{(n+1)^2}{n^2}\frac{2^n}{2^{n+1}}= \frac1{2} < 1,$$

and $\sum a_n$ converges. Hence $\sum n^2 x^n$ is uniformly convergent on $[-1/2,1/2]$.