Consider $x_1,x_2,x_3,....,x_n\in \mathbb{N}^+$

What is the upperbound and lowerbound of the following expression

$R=\frac{\sum_{i=1}^{n-1}(x_i + x_{i+1})}{\sum_{i=1}^{n}x_i}$

Here is my trail.

Neumarator is always more than the denominator in $R$ since we are counting $x_{i+1}$ two times for all $i$. So, $R$ will be always more than 1.

If $x_1=x_2=x_3=\ldots =x_n=k$, then


Since $n\geq 1$, $R<2$

So, the range of $R$ is


Is it correct?


Hint: $$R=\frac{\sum_{i=1}^{n-1}(x_i + x_{i+1})}{\sum_{i=1}^{n}x_i}$$ $$=\frac{(\sum_{i=1}^{n}x_i)-x_n}{\sum_{i=1}^{n}x_i}+\frac{\sum_{i=1}^{n}x_{i}-x_1}{\sum_{i=1}^{n}x_i}$$ $$=2-\frac{x_n+x_{1}}{\sum_{i=1}^{n}x_i}$$

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  • $\begingroup$ why $\frac{\sum_{i=1}^{n-1}x_i}{\sum_{i=1}^{n}x_i}=1$ in the second line? $\endgroup$ – medved Jul 21 '16 at 20:12
  • $\begingroup$ My mistake; Let me fix. $\endgroup$ – Paul Jul 21 '16 at 20:20

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