Uniform convergence of a series of functions (M-test)

Question

I'm requested to study the convergence (pointwise and uniform) of the series: $$ \sum_{n=1}^\infty \left( 1+\frac{x}{n^2} \right)^{n^3}, \; \; x \in \Bbb R $$ I managed to demonstrate the pointwise convergence for $(x \lt 0)$:

1. $\lim_{\, n \rightarrow \infty} a_n = 0 \; \; \text{iff} \; \; x \lt 0$
2. I've studied the absolute convergence (maybe I'm wrong?)

Now I'm in trouble to demonstrate the uniform convergence (I tried to use M-test). Someone have any hint for me?

Answer 1

If the convergence were uniform on any interval $(b,0)$ where $b < 0$ then we must have $f_n(x) = (1+x/n^2)^{n^3} \to 0$ uniformly.

However, this is not possible since $-n^{-1} \in (b,0)$ for all sufficiently large $n$ and

$$\lim_{n \to \infty}f_n(-n^{-1}) = \lim_{n \to \infty}\left(1 + \frac{-n^{-1}}{n^2}\right)^{n^3}= \lim_{n \to \infty} \left(1 - \frac{1}{n^3}\right)^{n^3}= e^{-1} \neq 0$$

Try to finish by proving uniform convergence on any interval $(b,a)$ where $b < a < 0$ using the Weierstrass M-test.

Hint: $\displaystyle \left( 1- \frac{x}{n}\right)^n \leqslant e^{-x}$ for all $0 \leqslant x \leqslant n$

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On proving non-uniform convergence of a series of functions

A necessary condition for the uniform convergence of a series of functions $\sum f_n(x)$ for $x \in D$ is that the sequence $(f_n(x))$ converges uniformly to $0$ for $x \in D$. This can be proved using the Cauchy criterion for uniform convergnece of a series.

We say that $f_n(x) \to 0$ uniformly for all $x \in D$ if for every $\epsilon > 0$ there exists $N \in \mathbb{N}$ which may depend on $\epsilon$ but not on $x$ such that if $n \geqslant N$ then $|f_n(x)| < \epsilon$ for all $x \in D$. Negating this statement we see that $f_n(x)$ does not converge uniformly if there exists $\epsilon_0 > 0$ such that for all $n$ there exists some $k \geqslant n$ and some $x_k \in D$ such that $|f_k(x_k)| > \epsilon_0$.

In this problem we have $f_n(x) = (1+x/n^2)^{n^3}$ and $D = (b,0)$. Taking $x_n = -n^{-1}$ we have

$$f_n(x_n) = \left(1 - \frac{1}{n^3}\right)^{n^3} \underset{n \to \infty}\longrightarrow e^{-1}$$

Hence, there exists $M$ such that $|f_n(x_n)|= f_n(x_n) > \frac{e^{-1}}{2}:= \epsilon_0$ for all $n \geqslant M$.

Going back to the condition for non-uniform convergence we have produced $\epsilon_0 = \frac{e^{-1}}{2}$ such that for any $n$ we can find $k = \max(n,M)$ and $x_k = -k^{-1}$ such that $|f_k(x_k) > \epsilon_0$. Hence, the condition for nnuniform convergence is fulfilled.