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Let $ \{\{ a_n\}_m\} $ be a sequence of converging sequences, such that $ \sum_{m=0}^{\infty}\{a_n\}_m $ converges absolutely for all $ n $.

Is it true that $ \lim_n \sum_{m=0}^{\infty} \{a_n\}_m = \sum_{m=0}^{\infty}\lim_n \{a_n\}_m $?

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please adjust the subscripts specially in the summation. – NoChance Nov 21 '11 at 12:34
up vote 5 down vote accepted

No (I understood the question differently from Gerry). Take $a_n(m) = 1/n$ if $m \lt n$ and $a_n(m) = 0$ otherwise (here $a_0(m) = 0$ is understood).

Then $\lim_m a_n(m) = 0$ for all $n$, hence each sequence $(a_n(m))_{m=0}^\infty$ converges.

On the other hand, $\sum_{m=0}^{\infty} a_n(m) = \sum_{m=0}^{n-1} \frac{1}{n} = 1$ for all $n \geq 1$ while $\lim_{n} a_n(m) = 0$ for all $m$ so that $\sum_{m=0}^\infty \lim_{n} a_n(m) = 0$.

Alternatively, take $a_{n}(m) = 1$ if $n=m$ and $a_{n}(m) = 0$ otherwise. Then the same properties as above hold.

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That doesn't hold as well, as I said the series $ \sum _{m=0} ^{\infty} {a_n(m)}$ converges for all $ n $, yet for $ n = 2 $ we get that $ \sum a_2 (m) = \sum 1/2 = \infty $ – Shai Deshe Nov 21 '11 at 12:47
I don't understand what you're saying. For $n=2$ I have $a_2(0) = 1/2$, $a_2(1) = 1/2$ and $a_2(m) = 0$ for $m \geq 2$. – t.b. Nov 21 '11 at 12:48
OK, now I get it – Shai Deshe Nov 21 '11 at 12:50

Perhaps an even simpler counterexample would be

$$ a_{n,m} = \begin{cases}1 & \text{if } n = m \\ 0 & \text{otherwise.}\end{cases} $$


  • $\displaystyle \lim_{n\to\infty} a_{n,m} = 0$ for all $m$,
  • $\displaystyle \sum_{m=0}^\infty |a_{n,m}| = \sum_{m=0}^\infty a_{n,m} = 1$ for all $n$,

and therefore

  • $\displaystyle \lim_{n\to\infty} \sum_{m=0}^\infty a_{n,m} = 1$, but
  • $\displaystyle \sum_{m=0}^\infty \lim_{n\to\infty} a_{n,m} = 0$.
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Oh, sorry, I didn't see your answer before adding the second example. – t.b. Nov 21 '11 at 12:54

The subscripts in the original are out of whack, so maybe I misunderstand, but if $a_n=1,0,0,0,0,\dots$ for even $m$, and $a_n=0,1,0,0,0,0,\dots$ for odd $m$, then the limit on the left is 1, but the sum on the right doesn't exist because $\lim_na_n$ doesn't exist.

EDIT: now that the problem statement has been edited into something that makes sense, I'll piggyback on the answers that have been posted and note that if we take $a_{n,m}=n$ if $m=n$ and $0$ otherwise, then the limit of the sums is infinite, while the sum of the limits is zero, so they can be about as different as you like. Or take $a_{n,m}=(-1)^n$ if $m=n$, $0$ otherwise, and the limit of the sums doesn't exist, while the sum of the limits is $0$.

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Generally true, yet I did mention that each of the sequences $ a_n $ do converge. – Shai Deshe Nov 21 '11 at 12:40

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