electrical engineers sometimes play fast-and-loose with some of the mathematics, particular from the P.O.V. of mathematicians.
consider the ideal reconstruction of this continuous-time signal from its samples:
$$ x(t) = \sum\limits_{n=-\infty}^{+\infty} x[n] \operatorname{sinc} \left(\tfrac1T(t-nT) \right) $$
it's not hard to see that $x(nT)=x[n]$. one might think that if finite samples $x[n]$ went into the reconstruction, a finite output $x(t)$ would come out. but this sequence will result in a spike of $\infty$ in height:
$$ x[n] = \begin{cases}
(-1)^{n+1} \qquad & n<0 \\
(-1)^n \qquad & n \ge 0 \\
\end{cases} $$
consider the value of $x(t)$ when $t=-\tfrac{T}2$:
$$\begin{align}
x(t) &= \sum\limits_{n=-\infty}^{+\infty} x[n] \operatorname{sinc} \left(\tfrac1T(t-nT) \right) \\
&= \sum\limits_{n=-\infty}^{-1} x[n] \operatorname{sinc} \left(\tfrac1T(t-nT) \right) + \sum\limits_{n=0}^{+\infty} x[n] \operatorname{sinc} \left(\tfrac1T(t-nT) \right) \\
&= \sum\limits_{n=-\infty}^{-1} (-1)^{n+1} \operatorname{sinc} \left(\tfrac1T(t-nT) \right) + \sum\limits_{n=0}^{+\infty} (-1)^n \operatorname{sinc} \left(\tfrac1T(t-nT) \right) \\
&= \sum\limits_{n=-\infty}^{-1} -(-1)^n \operatorname{sinc} \left(\tfrac1T(t-nT) \right) + \sum\limits_{n=0}^{+\infty} (-1)^n \operatorname{sinc} \left(\tfrac1T(t-nT) \right) \\
&= \sum\limits_{n=1}^{+\infty} -(-1)^{-n} \operatorname{sinc} \left(\tfrac1T(t-(-n)T) \right) + \sum\limits_{n=0}^{+\infty} (-1)^n \operatorname{sinc} \left(\tfrac1T(t-nT) \right) \\
&= -\sum\limits_{n=1}^{+\infty} (-1)^n \operatorname{sinc} \left(\tfrac1T(t+nT) \right) + \sum\limits_{n=0}^{+\infty} (-1)^n \operatorname{sinc} \left(\tfrac1T(t-nT) \right) \\
&= \operatorname{sinc} \left(\tfrac{t}T \right) + \sum\limits_{n=1}^{+\infty} (-1)^n \left( \operatorname{sinc}\left(\tfrac1T(t-nT) \right) - \operatorname{sinc}\left(\tfrac1T(t+nT) \right)\right) \\
\end{align}$$
now when $t=-\tfrac{T}2$:
$$\begin{align}
x\left(-\tfrac{T}2 \right) &= \operatorname{sinc} \left(\tfrac12 \right) + \sum\limits_{n=1}^{+\infty} (-1)^n \Big( \operatorname{sinc}\left(-\tfrac12-n \right) - \operatorname{sinc}\left(-\tfrac12+n \right) \Big) \\
&= \operatorname{sinc} \left(\tfrac12 \right) + \sum\limits_{n=1}^{+\infty} (-1)^n \Big( \operatorname{sinc}\left(n+\tfrac12\right) - \operatorname{sinc}\left(n-\tfrac12\right) \Big) \\
&= \operatorname{sinc} \left(\tfrac12 \right) + \sum\limits_{n=1}^{+\infty} (-1)^n \left( \frac{\sin\left(\pi n+\tfrac\pi 2\right)}{\pi n+\tfrac\pi 2} - \frac{\sin\left(\pi n-\tfrac\pi 2\right)}{\pi n-\tfrac\pi 2} \right) \\
&= \operatorname{sinc} \left(\tfrac12 \right) + \sum\limits_{n=1}^{+\infty} (-1)^n \left( \frac{\cos\left(\pi n\right)}{\pi n+\tfrac\pi 2} - \frac{-\cos\left(\pi n\right)}{\pi n-\tfrac\pi 2} \right) \\
&= \operatorname{sinc} \left(\tfrac12 \right) + \sum\limits_{n=1}^{+\infty} (-1)^n \cos\left(\pi n\right) \left( \frac{1}{\pi n+\tfrac\pi 2} + \frac{1}{\pi n-\tfrac\pi 2} \right) \\
&= \operatorname{sinc} \left(\tfrac12 \right) + \sum\limits_{n=1}^{+\infty} (-1)^n (-1)^n \left( \frac{1}{\pi n+\tfrac\pi 2} + \frac{1}{\pi n-\tfrac\pi 2} \right) \\
&= \operatorname{sinc} \left(\tfrac12 \right) + \sum\limits_{n=1}^{+\infty} \frac{2 \pi n}{(\pi n)^2 - (\tfrac\pi 2)^2} \\
& > \operatorname{sinc} \left(\tfrac12 \right) + \sum\limits_{n=1}^{+\infty} \frac{2 \pi n}{(\pi n)^2 } \\
& = \operatorname{sinc} \left(\tfrac12 \right) + \frac{2}{\pi}\sum\limits_{n=1}^{+\infty} \frac{1}{n} \to \infty \\
\end{align}$$
this is the output of a filter with impulse response
$$ h(t) = \tfrac1{T} \operatorname{sinc} \left( \tfrac{t}{T} \right) $$
and input
$$ x_\mathrm{s}(t) = T \sum\limits_{n=-\infty}^{+\infty} x[n] \delta(t-nT) $$
now you might rightfully say that the input (with those dirac impulse functions) is not bounded. that is correct.
so now consider this other input that is bounded:
$$ x_\mathrm{r}(t) = \sum\limits_{n=-\infty}^{+\infty} x[n] \operatorname{rect}\left(\tfrac1{T}(t-nT)\right) $$
where $\operatorname{rect}(u) \triangleq \begin{cases}
1 \qquad & |u|<\tfrac12 \\
\tfrac12 \qquad & |u|=\tfrac12 \\
0 \qquad & |u|>\tfrac12 \\
\end{cases} $
with $ x[n] = \begin{cases}
(-1)^{n+1} \qquad & n<0 \\
(-1)^n \qquad & n \ge 0 \\
\end{cases} $,
then $x_\mathrm{r}(t)$ is bounded
$$ |x_\mathrm{r}(t)| \le 1 \qquad \forall t \in \mathbb{R} $$
now do you thing that infinite sum will come out differently? (consider the commutative and associative properties of the multiplication operation. then consider of those properties should apply to convolution.)