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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

[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?

[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?

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