Check the continuity of the next function $f(x)=\sum_{n=1}^{\infty}(x+\frac{1}{n^2})^n$

Question

Check the continuity of the next function $f(x)=\sum_{n=1}^{\infty}(x+\frac{1}{n^2})^n$.

I've started by doing Cauchy test to check when the sum converges: $f(x)=\lim_{n\to\infty}\sqrt[n]{(x+\frac{1}{n^2})^n}=x$.

Now we know that the sum should converges whenever $x<1$.

So, Now I need to check uniform convergence. But I'm getting stuck on: $|f_n(x)-f(x)|=|(x+\frac{1}{n^2})^n-x|$.

What can I do?

Answer 1

As you stated by using the Cauchy test the series is convergent for $x\in(-1,1)$.

Now let $0<a<1$ then we have $$\left|x+\frac{1}{n^2}\right|^n\leq\left(a+\frac{1}{n^2}\right)^n=v_n\quad\forall x\in[-a,a]$$ and since $$v_n=\exp\left(n\log\left(a+\frac{1}{n^2}\right)\right)\sim_\infty a^n$$ then the series $\displaystyle\sum v_n$ is convergent by comparison with geometric convergent series. We deduce that the given series is normal convergent on every interval $[-a,a]\subset (-1,1)$ and then the function $f$ is continuous on $(-1,1)$.