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.

  • $\begingroup$ Are you looking for an example for which both expressions are finite? $\endgroup$ – Servaes Jan 10 '18 at 17:31
  • $\begingroup$ @Servaes that would be ideal, yes! $\endgroup$ – Pierpaolo Vivo Jan 10 '18 at 17:32
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    $\begingroup$ This works by using the standard sum $\sum_{k=1}^{n}1/n=1$. Just take $f(x, y) =1/y$ $\endgroup$ – Paramanand Singh Jan 11 '18 at 6:22
  • $\begingroup$ It's not the simplest, but one important example is the Eisenstein series $G_{2k}$. (It's dealing with exchanging two sums rather than a sum and a limit, but the principle is the same). For $2k \geq 4$, the sum invovled converges uniformly, and the sums can be freely exchanged. For $2k = 2$ it fails, and extra term pops out that ruins the behavior under the modular group. (I mention it because a lot of students think that issues like this are pedantry or simply methods that are correct but verboten because they haven't been covered in class.) $\endgroup$ – anomaly Jan 11 '18 at 19:30

Using Iverson brackets, $$ f(k,M)=[k=M] $$ $$ %f(k,M)=\left\{\begin{array}{} %0&\text{if }k\ne M\\ %1&\text{if }k=M %\end{array}\right. $$

  • $\begingroup$ Nice. Are the limits 1 and 0? $\endgroup$ – Eric Duminil Jan 11 '18 at 1:23
  • $\begingroup$ @EricDuminil: indeed, the limits are $1$ and $0$. $\endgroup$ – robjohn Jan 11 '18 at 4:36
  • $\begingroup$ In fact, you can get any two limits $a$ and $b$ you want by this method: $f(k, M) = 0$ for $M < k$, $a$ for $M = k$, and $b$ for $M > k$. $\endgroup$ – Michael Seifert Jan 11 '18 at 15:30
  • $\begingroup$ @MichaelSeifert: the sum of the limits is infinite for that unless $b=0$, is it not? $\endgroup$ – robjohn Jan 11 '18 at 15:37
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    $\begingroup$ @MichaelSeifert: $f(k,M)=a2^{-k}+(b-a)[k=M]$ or something like that would work. $\endgroup$ – robjohn Jan 11 '18 at 16:45

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,

$\lim_{M \rightarrow \infty} \frac{M(M+1)}{2M} = \lim_{M\rightarrow \infty} \frac{M+1}{2}$


$\sum^\infty_{t=1} \lim_{M\rightarrow \infty} \frac{t}{M} = 0 $,

as every summand is zero for every finite $t$.

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    $\begingroup$ I so far like this one the most. It's easy, well-known and natural to people new in calculus. $\endgroup$ – yo' Jan 11 '18 at 22:40

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.


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 $\sum_{t=1}^\infty\lim_{M\to\infty}1/M=0$.


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