convergence of sequences

[SemperFi]

How can I prove, that if (a_n) converges, (A_n) also converges, and ?
Thank you very much!


My idea was that , and I know that (a_k-a) < eps, so maybe ... Does that work?

 

[HallsofIvy]

You do not say that \(\displaystyle a= \sum a_n\) although that appears to be what you mean.

 

[Ishuda]

View attachment 4601

How can I prove, that if (a_n) converges, (A_n) also converges, and View attachment 4602?
Thank you very much!

Saw something a long time ago which reminded me of that. Had to go look it up, just couldn't remember the name of it - Cesàro summation [in this case the a_k would themselves be a partial sum and if those partial sums converged, the Cesàro sum converged to the same thing].

Just talking through an outline of the proof, without an actual proof, seems like we would have something like, if a_k converges to a then
A_n = \(\displaystyle \frac{1}{n} \Sigma_{k=1}^{k=n} a_k\) ~ \(\displaystyle \frac{1}{n} \Sigma_{k=1}^{k=m} a_k + \frac{n-m}{n} a = \frac{1}{n} \Sigma_{k=1}^{k=m} a_k - \frac{m}{n} a + a\)
or, as n goes to infinity, that fixed summation up to m which is divided by n and the m/n term go to zero so that A_n goes to a

Now if one could put that on a firm foundation, seems like they would have a proof.

 

[Ishuda]

... Does that work?

Not quite but almost. Consider the sequence
{a_k}={10000, 1, 1, 1, 1, 1, ...}
Not all |a_k - a| are less than epsilon. However there is some m for which that is true for all k > m and, given epsilon, m can be fixed.

 

[SemperFi]

Thank you very much Ishuda!!!