Does convergence of integral series and second derivative series imply convergence? Let $f_n : \mathbb R \to \mathbb [0,\infty)$ be a sequence of smooth functions such that
$$\sum_{n=1}^\infty \int_{\mathbb R} f_n(x) dx$$
is convergent. Further, we have that
$$\sum_{n=1}^\infty f_n''(x)$$
converges for every $x \in \mathbb R$.
Does this imply that
$$\sum_{n=1}^\infty f_n(x)$$
converges for all $x\in\mathbb R$?
 A: No.
Consider $f_n(x) = \exp(-g(n) x^4)$ for a positive function $g : \Bbb N \to \Bbb R_+$ to be fixed later. (The functions are clearly smooth.)
To begin with, one can see that $\sum_{n = 1}^\infty f_n(0) = \sum_{n = 1}^\infty 1$ does not converge.
Let $c := \int_{\Bbb R}\exp(-x^4) \ {\mathrm d}x$. It is easy to see that this integral does converge.
Then, $$\int_{\Bbb R} f_n = \frac{c}{\sqrt[4]{g(n)}},$$
as seen by an easy change of variables. (Which works since $g > 0$.)
On the other hand, one can compute the double derivative as $$f_n''(x) = [(4g(n))^2 x^6 \exp(-g(n)x^4)] - [12 g(n) x^2 \exp(-g(n)x^4)] \tag{1}.$$
The task now is to fix $g$ in such a way that both the above expressions converge as we sum over all $n \geqslant 1$. A polynomial expression seems to be hopeful and it does work.
More precisely, consider $g(n) := n^8$. Then, the sum of integrals converges since $\sum \frac{1}{n^2} < \infty$.
To see the same for $\sum_{n = 0}^\infty f''_n(x)$, first note $f_n''(0) = 0$ for all $n$.
Now, if $x \neq 0$, then one can note that both the terms in $[\cdots]$ in $(1)$ have a convergent sum by an application of the ratio test. Indeed, one only needs to check that
$$\lim_{n \to \infty} \frac{(n + 1)^{k}}{n^{k}}\frac{\exp(-(n + 1)^8 x^4)}{\exp(-n^8 x^4)} = 0$$
for $k = 8, 16$.
