# [SOLVED]Problem involving m-tail of a sequence

#### issacnewton

##### Member
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
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
I think that should be $$\displaystyle X_m Y_m$$ .

#### issacnewton

##### Member
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
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
Hello Makarov,

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