Can this sequence equation be proven?

[Miss_lolitta]

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

If [math]
a_n,b_n
[/math]are sequences of real number ,n>m then:

[math]
a_{n+1}S_n-a_m S_{m-1}+\sum_{k=m}^{n}( a_k - b_{k+1})S_k
[/math]
Where [math]
S_n
[/math]is the partial sum of sequence [math]
\sum_{k=1}^{\infty}b_n
[/math]

Thanks for any help

 

[matt grime]

the tag here is tex, not math, in the square brackets.

If <br /> a_n,b_n<br />
are sequences of real number ,n>m then:

<br /> a_{n+1}S_n-a_m S_{m-1}+\sum_{k=m}^{n}( a_k - b_{k+1})S_k<br />
Where <br /> S_n<br />is the partial sum of sequence <br /> \sum_{k=1}^{\infty}b_n<br />

nope, still makes no sense.

 

[Miss_lolitta]

yes that's right

thanks

 

[mathwonk]

what color is your hair?

 

[Miss_lolitta]

silver

 

[HallsofIvy]

If <br /> a_n,b_n<br />
are sequences of real number ,n>m then:

<br /> a_{n+1}S_n-a_m S_{m-1}+\sum_{k=m}^{n}( a_k - b_{k+1})S_k<br />

equals what??

Where <br /> S_n<br />is the partial sum of sequence <br /> \sum_{k=1}^{\infty}b_n<br />

Presumably you mean "Where <br /> S_n<br />is the partial sum of sequence
\sum_{k=1}^n b_n

 

[Miss_lolitta]

sorry

\sum_{k=1}^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