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

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

nope, still makes no sense.

 

[Miss_lolitta]

yes that's right

thanks

 

[mathwonk]

what color is your hair?

 

[Miss_lolitta]

silver

 

[HallsofIvy]

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

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

equals what??

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

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

 

[Miss_lolitta]

sorry

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