Uniform convergence of $\sum\limits_{n=1}^∞n^{-x}(e^{\frac{x}{n^2}}-1)$ 
Pointwise and uniform convergence of the following series of functions:
  $$\sum_{n=1}^{\infty} n^{-x}\left(e^{\frac{x}{n^2}}-1\right).$$

Now, the series of function converges pointwise as $x \in (-1, +\infty)$, because, as $n \to \infty$, $$f_n(x)=\mathcal{O}\left(\frac{x}{n^{x+2}}\right). \quad \forall x\in \mathbb{R}$$
I am having some problems with the uniform convergence because I could not find the supremum. Any suggestions?
 A: $\def\e{\mathrm{e}}\def\peq{\mathrm{\phantom{=}}{}}$First, since$$
\frac{1}{n^x} \left(\exp\left( \frac{x}{n^2} \right) - 1 \right) \sim \frac{x}{n^{x + 2}}\quad (n → ∞)
$$
for any fixed $x$, then $\displaystyle \sum_{n = 1}^∞ \frac{1}{n^x} \left( \exp\left( \frac{x}{n^2} \right) - 1 \right)$ converges iff $x + 2 > 1$, i.e. $x \in (-1, +∞)$.
Next, to prove that the series does not converge uniformly for $x \in (-1, +∞)$, it suffices to prove the following lemma.

Lemma: Given that $\{u_n(x)\} \subset C([a, b])$. If $\displaystyle \sum_{n = 1}^∞ u_n(x)$ converges pointwise for $x \in (a, b]$, but $\displaystyle \sum_{n = 1}^∞ u_n(a)$ diverges, then $\displaystyle \sum_{n = 1}^∞ u_n(x)$ does not converge uniformly for $x \in (a, b]$.

Proof: Suppose that $\displaystyle \sum_{n = 1}^∞ u_n(x)$ converges uniformly for $x \in (a, b]$. For any fixed $ε > 0$, there exists $N \geqslant 1$ such that for any $m > n \geqslant N$,$$
\left| \sum_{k = n + 1}^m u_k(x) \right| = \left| \sum_{k = 1}^m u_k(x) - \sum_{k = 1}^n u_k(x) \right| < ε, \quad \forall x \in (a, b]
$$
which implies$$
\left| \sum_{k = n + 1}^m u_k(a) \right| = \lim_{x → a^+} \left| \sum_{k = n + 1}^m u_k(x) \right| \leqslant ε.
$$
Thus $\displaystyle \sum_{n = 1}^∞ u_n(a)$ converges, a contradiction.
Back to the question. Since the series diverges for $x = -1$, then it does not converge uniformly for $x \in (-1, 0]$ by the lemma, and thus not for $x \in (-1, +∞)$.
Finally, it will be proved that for any $δ > -1$, this series converges uniformly for $x \in [δ, +∞)$. It suffices to prove for $-1 < δ < 0$.
For $x \in [0, 2]$ and $n \geqslant 2$, note that$$
e^{-t} - 1 \geqslant -t \Longrightarrow \e^t - 1 \leqslant t\e^t, \quad \forall t \in \mathbb{R}
$$
thus\begin{align*}
&\peq \sum_{k = n}^∞ \left| \frac{1}{k^x} \left( \exp\left( \frac{x}{k^2} \right) - 1 \right) \right| = \sum_{k = n}^∞ \frac{1}{k^x} \left( \exp\left( \frac{x}{k^2} \right) - 1 \right)\\
&\leqslant \sum_{k = n}^∞ \frac{1}{k^x} · \frac{x}{k^2} \exp\left( \frac{x}{k^2} \right)= \sum_{k = n}^∞ \frac{x}{k^2} · \left( \frac{1}{k} \exp\left( \frac{1}{k^2} \right) \right)^x\\
&\leqslant \sum_{k = n}^∞ \frac{x}{k^2} · \left( \frac{1}{2} \exp\left( \frac{1}{4} \right) \right)^x \leqslant \sum_{k = n}^∞ \frac{x}{k^2} \leqslant \sum_{k = n}^∞ \frac{2}{k^2}.
\end{align*}
Since $\sum\limits_{n = 2}^∞ \dfrac{2}{n^2} < +∞$, then the given series converges uniformly for $x \in [0, 2]$.
For $x \in (2, +∞)$ and $n \geqslant 2$, note that $\dfrac{1}{k} \exp\left( \dfrac{1}{k^2} \right) \leqslant \dfrac{1}{2} \exp\left( \dfrac{1}{4} \right) < 1$ for $k \geqslant 2$, thus\begin{align*}
&\peq \sum_{k = n}^∞ \left| \frac{1}{k^x} \left( \exp\left( \frac{x}{k^2} \right) - 1 \right) \right| = \sum_{k = n}^∞ \frac{1}{k^x} \left( \exp\left( \frac{x}{k^2} \right) - 1 \right)\\
&\leqslant \sum_{k = n}^∞ \frac{1}{k^x} \exp\left( \frac{x}{k^2} \right) = \sum_{k = n}^∞ \left( \frac{1}{k} \exp\left( \frac{1}{k^2} \right) \right)^x \leqslant \sum_{k = n}^∞ \left( \frac{1}{k} \exp\left( \frac{1}{k^2} \right) \right)^2.
\end{align*}
Because$$
\sum_{n = 2}^∞ \left( \frac{1}{n} \exp\left( \frac{1}{n^2} \right) \right)^2 = \sum_{n = 2}^∞ \frac{1}{n^2} \left( \exp\left( \frac{2}{n^2} \right) - 1 \right) + \sum_{n = 2}^∞ \frac{1}{n^2} < +∞,
$$
then the given series converges uniformly for $x \in (2, +∞)$.
For $x \in [δ, 0)$ and $n \geqslant 1$, since$$
\sum_{k = n}^∞ \left| \frac{1}{k^x} \left( \exp\left( \frac{x}{k^2} \right) - 1 \right) \right| = \sum_{k = n}^∞ \frac{1}{k^x} \left( 1 - \exp\left( \frac{x}{k^2} \right) \right) \leqslant \sum_{k = n}^∞ \frac{1}{k^δ} \left( 1 - \exp\left( \frac{δ}{k^2} \right) \right),
$$
and $\displaystyle \sum_{n = 1}^∞ \frac{1}{n^δ} \left( 1 - \exp\left( \frac{δ}{n^2} \right) \right) < +∞$, then the given series converges uniformly for $x \in [δ, 0)$.
Therefore for any $ε > 0$, there exists $N_1, N_2, N_2 \geqslant 1$ such that\begin{gather*}
\sum_{k = n}^∞ \left| \frac{1}{k^x} \left( \exp\left( \frac{x}{k^2} \right) - 1 \right) \right| < ε, \quad \forall n \geqslant N_1,\ x \in [δ, 0)\\
\sum_{k = n}^∞ \left| \frac{1}{k^x} \left( \exp\left( \frac{x}{k^2} \right) - 1 \right) \right| < ε, \quad \forall n \geqslant N_2,\ x \in [0, 2]\\
\sum_{k = n}^∞ \left| \frac{1}{k^x} \left( \exp\left( \frac{x}{k^2} \right) - 1 \right) \right| < ε. \quad \forall n \geqslant N_3,\ x \in (2, +∞)
\end{gather*}
Taking $N = \max(N_1, N_2, N_3)$ yields$$
\sum_{k = n}^∞ \left| \frac{1}{k^x} \left( \exp\left( \frac{x}{k^2} \right) - 1 \right) \right| < ε. \quad \forall n \geqslant N,\ x \in [δ, +∞)
$$
Thus the given series converges uniformly for $x \in [δ, +∞)$.
