# Limit Theorems, Sequences

• Mar 26th 2009, 12:38 PM
bearej50
Limit Theorems, Sequences
Suppose that (sn) and (tn) are convergent sequences with lim sn = s and lim tn = t. Then lim (sntn) = st.

Write a proof to this theorem that does not use the theorem that states that "every convergent sequence is bounded." I think that this can be done by using the identity sntn - st = (sn - s)(tn - t) + s(tn - t) + t(sn - s). I am lost from here.
• Mar 26th 2009, 01:02 PM
Plato
If $\displaystyle \varepsilon > 0$ then choose $\displaystyle \delta = \min \left\{ {\sqrt {\frac{\varepsilon }{3}} ,\frac{\varepsilon } {{3\left( {\left| s \right| + 1} \right)}},\frac{\varepsilon } {{3\left( {\left| t \right| + 1} \right)}}} \right\}$.

Make $\displaystyle \left| {s_n - s} \right| < \delta \;\& \,\left| {t_n - t} \right| < \delta$.

Then use the identity.
• Mar 27th 2009, 06:33 AM
bearej50
I'm still lost...
• Mar 27th 2009, 04:09 PM
xalk
Quote:

Originally Posted by bearej50
I'm still lost...

The identities that you must use are:

1)$\displaystyle s_{n}t_{n}-st = s_{n}t_{n}-st_{n}+st_{n}-st = t_{n}(s_{n}-s)+s(t_{n}-t)$

OR

2) $\displaystyle s_{n}t_{n}-ts_{n}+ts_{n}-st = s_{n}(t_{n}-t)+t(s_{n}-s)$.

In using for example the (1) identity we can work as follows:

we want to find out if there exists a No ,k, belonging to the natural Nos,such that for any given positive No ε>0

$\displaystyle |s_{n}t_{n}-st| = |s_{n}t_{n}-st_{n}+st_{n}-st| = |t_{n}(s_{n}-s)+s(t_{n}-t)|<\epsilon$,whenever ,$\displaystyle n\geq k$.

WHERE n is any natural No.

But since $\displaystyle |t_{n}(s_{n}-s)+s(t_{n}-t)|\leq |t_{n}||s_{n}-s| + |s||t_{n}-t|\leq |t_{n}||s_{n}-s| + (|s|+1)|t_{n}-t|$,if we can find a ,k, such that for :

$\displaystyle n\geq k$ ,then $\displaystyle |t_{n}||s_{n}-s|<\frac{\epsilon}{2}$ ,and

$\displaystyle (|s|+1)|t_{n}-t|<\frac{\epsilon}{2}$,we will have proved the desired result.

Note instead of using |s| we use ||s|+1| in the above inequalities is to avoid the case where s=0.

HINT: In trying to find this ,k, it must be the max{$\displaystyle n_{1},n_{2},n_{3}$} i,e the maximum of three other natural Nos
• Mar 30th 2009, 04:21 PM
xalk
bearej50.

Since there is no respond from you ,i will assume that you still do not know how to do the limit,so i will prove it.

So we want to prove $\displaystyle \lim_{n\rightarrow\infty}{s_{n}t_{n}} = st$

Let ε>0.

Thus $\displaystyle \frac{\epsilon}{2(|t|+\epsilon)}>0$.................................................. ................................1

and $\displaystyle \frac{\epsilon}{2(|s| +1)}>0$.................................................. ........................................2

Since now $\displaystyle \lim_{n\rightarrow\infty}{s_{n}} = s$ and $\displaystyle \lim_{n\rightarrow\infty}{t_{n}} = t$,given any +ve No and thus even the +ve Nos (1),(2),ε,
we can find Natural Nos $\displaystyle n_{1},n_{2},n_{3}$ and such that:

If $\displaystyle n\geq n_{1}$,then $\displaystyle |t_{n}|<|t|+\epsilon$.................................................. ...........................................3

If $\displaystyle n\geq n_{2}$,then $\displaystyle |s_{n}-s|<\frac{\epsilon}{2(|t|+\epsilon)}$.................................................. .........................................4

If $\displaystyle n\geq n_{3}$,then $\displaystyle |t_{n}-t|<\frac{\epsilon}{2(|s|+1)}$.................................................. ...........................................5

Choose k = max{$\displaystyle n_{1},n_{2},n_{3}$}

Hence ,

If $\displaystyle n\geq k$,then $\displaystyle n\geq n_{1},n\geq n_{2},n\geq n_{3}$,

HENCE

$\displaystyle |t_{n}|<|t| +\epsilon$.................................................. ........................................6

$\displaystyle |s_{n}-s|<\frac{\epsilon}{2(|t|+\epsilon)}$.................................................. ........................................7

$\displaystyle |t_{n}-t|<\frac{\epsilon}{2(|s|+1)}$.................................................. .........................................8

Now multiply (6) by (7) and we get:

$\displaystyle |t_{n}||s_{n}-s|< (|t|+\epsilon)\frac{\epsilon}{2(|t|+\epsilon)} = \frac{\epsilon}{2}$.................................................. ...........................................9

And by multiplying (8) by (|s|+1) we get:

$\displaystyle (|s|+1)|t_{n}-t|< (|s|+1)\frac{\epsilon}{2(|s|+1)} =\frac{\epsilon}{2}$.................................................. ..........................................10

Add (9) and (10) we get:

$\displaystyle |t_{n}||s_{n}-s|+(|s|+1)|t_{n}-t|< \frac{\epsilon}{2} +\frac{\epsilon}{2} = \epsilon$.

Hence $\displaystyle |t_{n}s_{n}-ts|<\epsilon$.

Therefor we have proved ,given any +ve No ,ε Τhere exist a natural No k such that :

If $\displaystyle n\geq k$ then $\displaystyle |t_{n}s_{n}-ts|< \epsilon$ ,for all n
• Mar 30th 2009, 05:05 PM
Plato
Re the last post.
Did you note that the OP said not to use the bounded theorem ?