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Sir, I have three infinite summation

$A =J_1 \sum_{n=2}^\infty (n-1) f(n-2) \tag 1$ , $B =\sum_{n=0}^\infty f(n) \tag 2$ and $C =J_2\sum_{n=1}^\infty f(n-1) \tag 3$, with $f(0)=2,f(1)=5$ , $J_1$ and $J_2$ are constants

Question

Can we express $A $ as functions of $B $ and $C $ if possible? Means can we rewrite $A $ using $B $ and $C $ only

Note

Hint is that $\sum_{n=0}^\infty f(n) $ is a constant called $\psi$ but we are not aware of the value of it. It implies derivative of $\sum_{n=0}^\infty f(n) $ is $0$ only . Thanks

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  • $\begingroup$ Note that $J_2\sum_{n=2}^{\infty}(n-1)f(n-2)$ is not a function of $n$. $\endgroup$
    – JimmyK4542
    Commented Aug 5, 2014 at 3:57
  • $\begingroup$ Typo error..I have edited now $\endgroup$
    – Nirvana
    Commented Aug 5, 2014 at 3:58
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    $\begingroup$ If you shift indices, you can see that $C = J_2B$, but you can't write $A$ in terms of $B$ and $C$. $\endgroup$
    – JimmyK4542
    Commented Aug 5, 2014 at 4:00
  • $\begingroup$ Yes that is obvious in the case of C.. Looking for A actually.. So are you sure,it is an impossible task so that I should not waste time on this? $\endgroup$
    – Nirvana
    Commented Aug 5, 2014 at 4:01

1 Answer 1

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If you write $C$ in full, it is $J_2(f(0)+f(1)+f(2)+...)$, which is the same as $J_2B$.

If you write $A$ in full, it is $J_1(f(0)+2f(1)+3f(2)+...)$.

Now, if I increase $f(3)$ by 1, and decrease $f(4)$ by 1, then the sum $B$ remains the same, but $A$ changes by $+3-4=-1$. So knowing $B$ (and C) is not enough to know $A$.

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  • $\begingroup$ Let me put in this way.. Is there any way we can rewrite $ \sum_{n=0}^\infty n f(n) $ such that we can separate that n from f(n) $\endgroup$
    – Nirvana
    Commented Aug 5, 2014 at 4:05
  • $\begingroup$ Perhaps $\sum_{k=0}^{\infty}\sum_{n=k}^{\infty}f(n)$ is not what you want. I showed that two different sequences with the same $\sum f(n)$ have different $\sum nf(n)$, so you need more information than $\sum f(n)$. $\endgroup$
    – Empy2
    Commented Aug 5, 2014 at 4:09

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