# Thread: can someone prove this

1. ## can someone prove this

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

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

$\displaystyle \sum_{k=m}^{n}a_k .b_k=$$\displaystyle a_{n+1}S_n-a_m S_{m-1}+\sum_{k=m}^{n}( a_k - b_{k+1})S_k Where \displaystyle S_n is the partial sum of sequence \displaystyle \sum_{k=1}^{\infty}b_n Thanks for any help 2. Originally Posted by miss_lolitta Hello, can someone prove this to me as. Any help would help save thier hair I have not torn out as yet. If \displaystyle a_n,b_n are sequences of real number ,n>m then: \displaystyle \sum_{k=m}^{n}a_k .b_k=$$\displaystyle a_{n+1}S_n-a_m S_{m-1}+\sum_{k=m}^{n}( a_k - b_{k+1})S_k$

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

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 $\displaystyle a_{n+1}$, while the RHS is dependent
on $\displaystyle a_{n+1}$.

RonL

3. the relation is correct..sure

4. Were is the proof??

5. Originally Posted by miss_lolitta
Were is the proof??
There is no proof because it cant be an identity.

RonL

6. How??

7. Originally Posted by miss_lolitta
How??
We seek to prove that for arbitary sequences $\displaystyle a_i,\ i=1 \dots$ and $\displaystyle b_i,\ i=1 \dots$,
and $\displaystyle S_n$ the sequence of partial sums of the $\displaystyle b_i$'s:

$\displaystyle \sum_{k=m}^{n}a_k .b_k=$$\displaystyle a_{n+1}S_n-a_m S_{m-1}+\sum_{k=m}^{n}( a_k - b_{k+1})S_k\ \ \dots(1). Rearranging: \displaystyle a_{n+1}=(-\sum_{k=m}^{n}a_k .b_k+$$\displaystyle 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 $\displaystyle a_{n+1}$, so
$\displaystyle a_{n+1}$ is deternined by the sequence $\displaystyle b_i,\ i=1\dots$
and $\displaystyle a_i,\ i=1 \dots n$, and so $\displaystyle a_i,\ i=1 \dots$ is not an arbitary sequence,
or if it is then $\displaystyle (1)$ is not an identity.

RonL