Construct $b_n$ so that $\sum_{n=0}^{\infty}b_n=B$

Question

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 $c_n$.

Answer 1

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 $\alpha=(C-B)/(C-A)$.

Answer 2

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 $B=(1-\lambda)A+ \lambda C$.

Answer 3

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 $c_n$.

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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 $\lim b_n = \lim(\frac {C-B}{C-A}a_n + \frac {B-A}{C-A}c_n) = \frac {C-B}{C-A}A + \frac {B-A}{C-A}C =\frac {AC-BA + BC-AC}{C-A} = \frac{BC-BA}{C-A} = B\frac {C-A}{C-A} = B$.

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