∑ C An =C ∑ An (n from 1 to infinity) ... why?

[ShengyaoLiang]

If ∑ C An (n from 1 to infinity) converges, and C in Real, then ∑ C An is convergent with :

∑ C An =C ∑ An (n from 1 to infinity) .. why?

 

[EnumaElish]

What exactly is your question?

 

[DeadWolfe]

Well if you know the rigorous definition of convergence of series than the property is easy to prove.

 

[sutupidmath]

write with s_n the partial sum of the first series that you are supposing that converges. We know that if it converges then the limit of its partial sum must exist and me finite. After that write with s'_n the partial sum of the second series, it obviously is equal to
s'_n=cs_n by taking the limit of this too, you can easily prove what you are looking for.
Do u understadn how to do it now?? This is only one way of doing it though.

 

[HallsofIvy]

Yes, that's correct. $\sum_{n=0}^\infty CA_n$ is defined as the limit of the partial sums: $\sum_{n=0}^\infty CA_n= \lim_{N\rightarrow\infty} \sum_{n=0}^N CA_n$. For each partial sum, the distributive law gives $\sum_{n=0}^N CA_n= C\sum_{n=0}^N A_n$ and, of course, $\lim_{N\rightarrow\infty} C\sum_{n=0}^N A_n= C\lim_{N\rightarrow\infty} C\sum_{n=0}^N A_n$.

 

[ShengyaoLiang]

thanks...hehe i did wrong...

 

[sutupidmath]

Yes, that's correct. $\sum_{n=0}^\infty CA_n$ is defined as the limit of the partial sums: $\sum_{n=0}^\infty CA_n= \lim_{N\rightarrow\infty} \sum_{n=0}^N CA_n$. For each partial sum, the distributive law gives $\sum_{n=0}^N CA_n= C\sum_{n=0}^N A_n$ and, of course, $\lim_{N\rightarrow\infty} C\sum_{n=0}^N A_n= C\lim_{N\rightarrow\infty} C\sum_{n=0}^N A_n= \sum_{n=0}^\infty A_n$.

yeah halls at the very end meant this:

[tex]C\lim_{N\rightarrow\infty} \sum_{n=0}^N A_n= C\sum_{n=0}^\infty A_n[/itex]

but i am sure it was a typo, latex thing.