Convergent Series Identities

[dtl42]

[SOLVED] Convergent Series Identities

Homework Statement

a) If c is a number and $$\sum a_{n}$$ from n=1 to infinity is convergent to L, show that $$\sum ca_{n}$$ from n=1 to infinity is convergent to cL, using the precise definition of a sequence.

b)If $$\sum a_{n}$$ from n=1 to infinity and $$\sum b_{n}$$ from n=1 to infinity are convergent to X and Y respectively, show that $$\sum b_{n}+a_{n}$$ from n=1 to infinity is convergent to X+Y.

Homework Equations

I personally thought these were identities, and have no idea how to approach them.

The Attempt at a Solution

a) Maybe $$\sum a_{n}$$ from n=1 to infinity = $$Lim (S_{n})$$ as n goes to infinity, has something to do with it

 

[scottie_000]

These proofs essentially rely on the definition of convergence for infinite series, i.e. that the mth partial sum converges to some limit L as m goes to infinity
Can you reformulate the question in terms of partial sums?

(Note that you can do what you'd expect to do with sums when they are only finite, but not necessarily when they are infinite)

 

[dtl42]

So I need to show that $$Lim (c*S_{n}) = c*L$$? Where $$S_{n}$$ is the nth partial sum?

 

[scottie_000]

So I need to show that $$Lim (c*S_{n}) = c*L$$? Where $$S_{n}$$ is the nth partial sum?

That's right. Do you know how to prove things about limits of sequences?
(The epsilon-delta definition of a limits for sequences, or any other definition)

 

[tiny-tim]

Maybe $$\sum a_{n}$$ from n=1 to infinity = $$Lim (S_{n})$$ as n goes to infinity, has something to do with it

HI dtl42!

Yes, ∑an and lim(Sn) are the same thing.

So start by writing out the definition of the statement "lim(Sn) = L".

Then it should be fairly obvious how to adjust it.

 

[dtl42]

The definition of $$Lim(S_{n})$$ is $$|S_{n}-L|<\epsilon\rightarrow \forall n>N$$ right? Can I multiply both sides of $$|S_{n}-L|<\epsilon$$ by c? But then there are two cases depending on the sign of c right?

 

[tiny-tim]

The definition of $$Lim(S_{n})$$ is $$|S_{n}-L|<\epsilon\rightarrow \forall n>N$$ right? Can I multiply both sides of $$|S_{n}-L|<\epsilon$$ by c? But then there are two cases depending on the sign of c right?

Use |ab| = |a| times |b|.

 

[dtl42]

So $$|cS_{n}-cL|<\epsilon*|c| \rightarrow \forall n>N$$, from this can I say the final statement?

 

[tiny-tim]

So $$|cS_{n}-cL|<\epsilon*|c| \rightarrow \forall n>N$$, from this can I say the final statement?

mmm … not exactly …

you see, you have to start with "given any ε > 0, there exists an N such that …" and end with "< ε", not "< ε|c|"

So you should begin "given any ε > 0, there exists an N such that … |Sn - L| < ε/|c| "

 

[dtl42]

Hmm, I can't really think of any way to get to that conclusion? You can't say it just because $$|c|>0$$... I'm kinda stuck.

 

[tiny-tim]

You can choose ε to be anything (> 0).

You can have one ε for Sn, and a different ε for cSn

In particular, you can choose the first one to be 1/|c| times the second one.

 

[dtl42]

Ok, so can I say let $$\epsilon_{1}=\frac{\epsilon_{2}}{|c|}$$ so $$|S_{n}-L|<\epsilon_{1}\rightarrow \frac{\epsilon_{2}}{|c|}$$, so $$|S_{n}-L|<\frac{\epsilon_{2}}{|c|} \rightarrow |cS_{n}-cL|<\epsilon_{2}$$?

 

[tiny-tim]

Woohoo! ​

(and you can just write ε instead of ε2)

… and nice LaTeX, btw! ​

 

[dtl42]

Ok, tiny-tim, thanks for all the help, I am going to mark this as solved