can someone prove this

**miss_lolitta** · April 25th 2006, 11:20 AM

Hello,
can someone prove this to me as.
Any help would help save thier hair I have not torn out as yet.

If
$a_n,b_n$
are sequences of real number ,n>m then:

$\sum_{k=m}^{n}a_k .b_k=$ $a_{n+1}S_n-a_m S_{m-1}+\sum_{k=m}^{n}( a_k - b_{k+1})S_k$

Where
$S_n$
is the partial sum of sequence
$\sum_{k=1}^{\infty}b_n$

Thanks for any help

**CaptainBlack** · April 25th 2006, 01:12 PM

Originally Posted by miss_lolitta

Hello,
can someone prove this to me as.
Any help would help save thier hair I have not torn out as yet.

If
$a_n,b_n$
are sequences of real number ,n>m then:

$\sum_{k=m}^{n}a_k .b_k=$ $a_{n+1}S_n-a_m S_{m-1}+\sum_{k=m}^{n}( a_k - b_{k+1})S_k$

Where
$S_n$
is the partial sum of sequence
$\sum_{k=1}^{\infty}b_n$

Thanks for any help

Either my correction of the TeX has introduced an error, or this cannot
be an identity, as the LHS is independent of $a_{n+1}$ , while the RHS is dependent
on $a_{n+1}$ .

RonL

**miss_lolitta** · April 27th 2006, 06:28 AM

the relation is correct..sure

**miss_lolitta** · April 29th 2006, 11:22 AM

Were is the proof??

**CaptainBlack** · April 29th 2006, 11:25 AM

Originally Posted by miss_lolitta

Were is the proof??

There is no proof because it cant be an identity.

RonL

**miss_lolitta** · April 29th 2006, 11:29 AM

How??

**CaptainBlack** · April 29th 2006, 12:25 PM

Originally Posted by miss_lolitta

How??

We seek to prove that for arbitary sequences $a_i,\ i=1 \dots$ and $b_i,\ i=1 \dots$ ,
and $S_n$ the sequence of partial sums of the $b_i$ 's:

$\sum_{k=m}^{n}a_k .b_k=$ $a_{n+1}S_n-a_m S_{m-1}+\sum_{k=m}^{n}( a_k - b_{k+1})S_k\ \ \dots(1)$ .

Rearranging:

$a_{n+1}=(-\sum_{k=m}^{n}a_k .b_k+$ $a_{n+1}S_n-a_m S_{m-1}+\sum_{k=m}^{n}( a_k - b_{k+1})S_k\left)/S_n$

Now everything on the RHS is independent of $a_{n+1}$ , so
$a_{n+1}$ is deternined by the sequence $b_i,\ i=1\dots$
and $a_i,\ i=1 \dots n$ , and so $a_i,\ i=1 \dots$ is not an arbitary sequence,
or if it is then $(1)$ is not an identity.

RonL

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