Find the simplest counterexample against exchanging limit and summation I would need a very simple counterexample to show that
$$
\lim_{M\to\infty}\sum_{t=1}^M f(t,M)
$$
may not necessarily be equal to
$$
\sum_{t=1}^\infty \lim_{M\to\infty}f(t,M)\ .
$$
The situation here is (slightly) different from the commonly asked question about interchanging limits and infinite summation, as $M$ is itself driving the upper limit of the sum. Can you exhibit a simple function $f$ which does the job? [Note that it should depend explicitly on $M$!]. I could only come up with an overly complicated situation, but I think I am missing something potentially very simple... Many thanks for you help. 
 A: Using Iverson brackets,
$$
f(k,M)=[k=M]
$$
$$
%f(k,M)=\left\{\begin{array}{}
%0&\text{if }k\ne M\\
%1&\text{if }k=M
%\end{array}\right.
$$
A: Let $f(t,M) =\frac{t}{M}$, then
$\lim_{M\rightarrow \infty} \sum_{t=1}^{M} \frac{t}{M} = \infty$
as it is just the arithmetic series over M,
$\lim_{M \rightarrow \infty} \frac{M(M+1)}{2M} = \lim_{M\rightarrow \infty} \frac{M+1}{2}$
while
$\sum^\infty_{t=1} \lim_{M\rightarrow \infty} \frac{t}{M} = 0 $,
as every summand is zero for every finite $t$.
A: One example I can think of is:  $f(t,M)=\frac1M g(\frac tM)$ where $g$ is continuous on $[0,1]$, so that $$\lim_{M\rightarrow \infty}\sum_{t=0}^{M}f(t,M)=\int_0^1 g(x)\,\mathrm{d}x$$
but $\sum_{t=0}^{\infty}\lim_{M\rightarrow \infty}f(t,M)=0$ if $g(0)\neq 0.$
For example you can choose $g(x)=x+1\text{ and }f(t,M)=\frac1M (1+\frac tM).$
EDIT: Of course that's a Riemann series, I thought I'd mention it.
A: Similar to @cgrudz's answer but a bit simpler. 
$$f(t,M)=1/M$$ 
Then $\lim_{M\to\infty}\sum_{t=1}^M1/M=1$ while $\sum_{t=1}^\infty\lim_{M\to\infty}1/M=0$.
