Welcome to our community

Be a part of something great, join today!

[SOLVED] Problem involving m-tail of a sequence

issacnewton

Member
Jan 30, 2012
61
Hello


I want to prove the following.
Let \(X\) and \(Y\) be two sequences,and \(XY\) converges. Then prove that
\(X_mY\) also converges,where
\[ X_m = \mbox{ m-tail of X } = (x_{m+n}\;:\; n\in \mathbb{N}) \]
Here is my proof.
let \(\lim\;(XY) = a \) . Then we have
\[ \forall \varepsilon >0\; \exists K_1 \in \mathbb{N} \;\forall n\geqslant K_1 \]
\[ |x_ny_n - a| < \varepsilon \]
Now let \(K = \max(K_1,\; m+1) \). The \( \forall n \geqslant K \) we have
\[|x_n y_n - a| < \varepsilon \].
But now all the \( (x_n) \) terms are values from the sequence \(X_m\). So
we have proven that
\[ \forall \varepsilon >0 \;\exists K \in \mathbb{N} \;\forall n\geqslant K \]
\[ |x_ny_n - a| < \varepsilon \]
where \( x_n\) values are from sequence \(X_m\). This proves that
\[ \lim(X_m Y) = a \]


which proves that \(X_mY\) also converges. Let me know if this is right
Thanks
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
I want to prove the following.
Let \(X\) and \(Y\) be two sequences, and \(XY\) converges. Then prove that
\(X_mY\) also converges,where
\[ X_m = \mbox{ m-tail of X } = (x_{m+n}\;:\; n\in \mathbb{N}) \]
This result is not true unless the sequences satisfy some additional conditions. For example, suppose that the sequences $X$ and $Y$ are given by $$X = (0,1,0,1,0,1,0,1,\ldots), \qquad Y = (1,0,2,0,3,0,4,0,\ldots).$$ Then the sequence $XY$ consists entirely of zeros, and therefore converges. But if $m$ is an odd number then the sequence $X_mY$ is the same as $Y$, and it diverges.
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
I think that should be \(\displaystyle X_m Y_m \) .
 

issacnewton

Member
Jan 30, 2012
61
Thanks Opalg

I see the mistake. But I am trying to see which step in my proof fails. I was trying to prove this result for using in another proof. And I just checked that another problem and it says that \(X\) does converge to some non zero number.

So which step fails in my proof.

Thanks
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
So
we have proven that
\[ \forall \varepsilon >0 \;\exists K \in \mathbb{N} \;\forall n\geqslant K \]
\[ |x_ny_n - a| < \varepsilon \]
where \( x_n\) values are from sequence \(X_m\). This proves that
\[ \lim(X_m Y) = a \]
The sequence $(x_ny_n)_{n=1}^\infty$ is $XY$ and not $X_mY$, assuming I interpret multiplication of sequences right. Then $X_mY$ is $(x_{m+n}y_n)_{n=1}^\infty$.
 

issacnewton

Member
Jan 30, 2012
61
Hello Makarov,

Like I said, I made mistake in stating the problem. Thanks for pointing that out.