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Say that I have a sequence $\{a_{n}\}_{n\in\mathbb{N}}$ $\subset$ [0,1] such that lim$_{N\rightarrow\infty}$ $\frac{1}{N}$ $\sum_{n=1}^{N}$ a$_{n}$ = a.

Can I somehow get the value of lim$_{(N-M)\rightarrow\infty}$ $\frac{1}{N-M}$ $\sum_{n=M}^{N-1}$ a$_{n}$ = a.

If that is not the case for general sequences, what specific conditions do the sequence need to satisfy for this? So far, only thing I could come up with was to try to manipulate one of them in hope that somehow the other limit would pop out. I have not been very successful yet.

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Put $K=N-M$ thus $N=M+K$ so $$\frac{1}{N-M}\sum_{n=M}^{N-1}a_n=\frac{1}{K}\sum_{n=M}^{M+K-1}a_n= \frac{1}{K}\sum_{k=1}^{K}a_{k-M+1}$$

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  • $\begingroup$ Thank you very much.. $\endgroup$ – HumbleStudent Oct 27 '19 at 21:02
  • $\begingroup$ Thank you..i corrected it. $\endgroup$ – Marios Gretsas Oct 27 '19 at 21:03

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