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There are given two real numbers, convergent series $\sum_{n=0}^{\infty}a_n=A$ and $\sum_{n=0}^{\infty}c_n=C$, such that $a_n<c_n\:\forall n\in\mathbb{N}$. Let $B\in(A,C)$; Construct $b_n$ such that $$\sum_{n=0}^{\infty}b_n=B$$ and $a_n<b_n<c_n\:\:\forall n\in\mathbb{N}$.

I can't figure out such $b_n$ because its hard to meet the last condition: $a_n<b_n<c_n\:\:\forall n\in\mathbb{N}$, I was trying with some convex combinations of $a_n$ and $c_n$.

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    $\begingroup$ If $b_n$ is a convex optimization of $a_n$ and $c_n$ then it is always the case that $a_n < b_n < c_n$. $\endgroup$
    – Leo
    Aug 19 '19 at 18:45
  • $\begingroup$ I know, thats why I was trying it but didnt come up with the solution. $\endgroup$ Aug 19 '19 at 18:46
  • $\begingroup$ I meant to say that your last sentence is unclear, as you say it was hard to meet the condition by doing something that, by definition, meets the condition. Anyway, others have already provided answers. $\endgroup$
    – Leo
    Aug 19 '19 at 18:53
  • $\begingroup$ It's very easy to meet $a_n < b_n < c_n$. Just set $b_n = \frac {a_n+c_n}2$. But then $\lim b_n = \lim \frac {a_n + c_n}2=\frac {A+C}2 $ which may not be $B$. $\endgroup$
    – fleablood
    Aug 19 '19 at 19:09
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You want $b_n=\alpha a_n+(1-\alpha)c_n$, where $\alpha\in(0,1)$, with $B=\alpha A+(1-\alpha)C$. Solving this last equation for $\alpha$ yields $\alpha=(C-B)/(C-A)$.

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If for all $n \in \mathbb N$ $0\le a_n < c_n$,

$$b_n =(1-\lambda)a_n + \lambda c_n$$ will work where $\lambda \in (0,1)$ is such that $B=(1-\lambda)A+ \lambda C$.

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Just scale them.

$A < B < C$ so $B$ is some proportion of the distance between $A$ and $C$. Just choose each $b_n$ so it is the same proportion between $a_n$ and $c_n$.

======== details ====

In other words $(B-A) = k*(C-A)$ for some $k; 0< k < 1$, so $k = \frac {B-A}{C-A}$ and $B = A + k*(C-A)= (1-k)A + kC$.

Let $b_n = (1-k)a_n + k{C-A}c_n= (1 - \frac{B-A}{C-A})a_n + \frac {B-A}{C-A}c_n = \frac {C-A -(B-A)}{C-A}a_n + \frac {B-A}{C-A}c_n= \frac {C-B}{C-A}a_n + \frac {B-A}{C-A}c_n$.

It should be clear thatn $\lim b_n = (1-k)\lim a_n + k\lim c_n = (1-k)A + kC=B$

Or, if you like, that $\lim b_n = \lim(\frac {C-B}{C-A}a_n + \frac {B-A}{C-A}c_n) = \frac {C-B}{C-A}A + \frac {B-A}{C-A}C =\frac {AC-BA + BC-AC}{C-A} = \frac{BC-BA}{C-A} = B\frac {C-A}{C-A} = B$.

.....

It really, really works no matter how you look at it. Linear scaling is linear scaling and it is pretty basic....

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