Tail of a Sequence

**veronicak5678** · August 13th 2011, 08:47 AM

Prove a sequence S_n converges iff its tail S_n+M converges.

**abhishekkgp** · August 13th 2011, 10:10 AM

Originally Posted by veronicak5678

Prove a sequence S_n converges iff its tail S_n+M converges.

do you mean $S_n$ converges iff $S_{n+M}$ converges?

**veronicak5678** · August 13th 2011, 10:45 AM

Yes, that's what I mean. Sorry I don't know Latex!
So... any ideas?

**abhishekkgp** · August 13th 2011, 07:51 PM

Originally Posted by veronicak5678

Yes, that's what I mean. Sorry I don't know Latex!
So... any ideas?

you just have to use the definition of convergence.

Assume $S_n \rightarrow a$ . we will show that $S_{n+M} \rightarrow a$ .

$S_n \rightarrow a$ gives us:
$\forall \epsilon >0, \exists N \text{ such that } n>N \Rightarrow |S_n-a|<\epsilon$

To prove $S_{n+M} \rightarrow a$ , that is:

$\forall \epsilon>0, \exists N^{'} \text{ such that } n>N^{'} \Rightarrow |S_{n+M}-a|<\epsilon$ . We easily see that $N^{'}=N-M$ does the job.

Now can you try and do the 'only if' part.

**veronicak5678** · August 13th 2011, 09:45 PM

I see what I was doing wrong. Thanks everyone.

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