= Solution
For integer $k\geq1$, define $f_k(n)=f(k,n)$. The assumed monotonicity in $t$ gives
$$
f_k(n)\uparrow\lim_{t\to\infty}f(t,n)
$$
for every $n$. Apply the <monotone convergence theorem> on $\mathbb N$ with counting measure:
$$
\lim_{k\to\infty}\sum_{n\geq1}f(k,n)
=\sum_{n\geq1}\lim_{k\to\infty}f(k,n)
=\sum_{n\geq1}\lim_{t\to\infty}f(t,n).
$$
The function $t\mapsto\sum_nf(t,n)$ is itself nondecreasing, and the positive integers are cofinal in $\mathbb R$ as $t\to\infty$, so its limit over real $t$ equals its limit over integer $k$. Therefore
$$
\boxed{
\lim_{t\to\infty}\sum_{n\geq1}f(t,n)
=\sum_{n\geq1}\lim_{t\to\infty}f(t,n)}.
$$
Without monotonicity, take
$$
f(t,n)=\mathbf1_{[n,n+1)}(t)
\qquad(t\geq1).
$$
For each fixed $n$, $f(t,n)\to0$, so the sum of the pointwise limits is zero. At every $t\geq1$, however, exactly one term is one, so $\sum_nf(t,n)=1$ and its limit is one. This <moving-spike obstruction to interchanging a limit and an infinite sum> gives the required counterexample.
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