I understand that the limit of the sum of two sequences equals the sum of the sequences' limis: [tex]\displaysyle \lim_{n\rightarrow\infty} (a_{n} + b_{n}) = \lim_{n\rightarrow\infty}a_{n} + \lim_{n\rightarrow\infty}b_{n}[/tex]. Similar results consequenly hold for sums of three sequences, four sequences, etc. (Call this the "finite distributive property.") But what about the sum of countably many sequences? Given the sequences [tex] \{a_{i} (n)\} [/tex], is it the case that(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\displaystyle \lim_{n\rightarrow\infty} \sum_{i=1}^{\infty} a_{i}(n) = \sum_{i=1}^{\infty} \lim_{n\rightarrow\infty}a_{i}(n) ?[/tex]

If not, are there any extra conditions on the sequences sufficient to make the equality hold? Any necessary and sufficient ones? I have an argument that if if both sides of the equality are finite, then the it holds. However, I've very likely made a mistake and would appreciate it if you guys could/would point out any errors.

Let [tex]\displaystyle L := \Sigma_{i=1}^{\infty} \lim_{n\rightarrow\infty}a_{i}(n)[/tex]. So [tex]\displaysyle \forall \epsilon > 0 \exists S \forall s \geq S \ |L - \Sigma_{i=1}^{s}\lim_{n\rightarrow\infty}a_{i}(n)| < \frac{\epsilon}{2}.[/tex] Also let [tex]L_{n} := \Sigma_{i=1}^{\infty}a_{i}(n)[/tex]. We now therefore have, given the same [tex]\frac{\epsilon}{2}[/tex] as before, [tex]\forall T \exists t \geq T \ |L_{n} - \Sigma_{i=1}^{t} a_{i}(n) | < \frac{\epsilon}{2}[/tex]. Taking n to the limit on both sides yields [tex]\displaystyle \lim_{n\rightarrow\infty} |L_{n} - \Sigma_{i=1}^{t} a_{i}(n) | = |\lim_{n\rightarrow\infty}L_{n} - \Sigma_{i=1}^{t} \lim_{n\rightarrow\infty} a_{i}(n) | \leq \frac{\epsilon}{2}[/tex] (by the finite distributive property). Finally, define [tex]L_{0} := \lim_{n\rightarrow\infty}L_{n}, \ U := \max(S, T)[/tex]. Then for each [tex] u \geq U [/tex], we have both [tex] |L - \Sigma_{i=1}^{u} \lim_{n\rightarrow\infty} a_{i}(n) | < \frac{\epsilon}{2}[/tex] and [tex]|L_{0} - \Sigma_{i=1}^{u} \lim_{n\rightarrow\infty} a_{i}(n) | \leq \frac{\epsilon}{2}[/tex]. By the triangle inequality, then, [tex] |L - L_{0}| < \epsilon[/tex]. Since epsilon was arbitrary, [tex]L = L_{0}[/tex].

What about the cases where the sides aren't finite?

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Limits of infinite sums of sequences

**Physics Forums | Science Articles, Homework Help, Discussion**