Double Sums by Dorin Marghidanu

Problem

Dorin Marghidanu sum

Solution

We have successively,

$\displaystyle \begin{align} \sum_{i=1}^n\left(a_i\cdot\sum_{k=1}^i\frac{1}{a_k}\right)&=a_1\sum_{k=1}^1\frac{1}{a_k}+a_2\sum_{k=1}^2\frac{1}{a_k}+\ldots+a_n\sum_{k=1}^n\frac{1}{a_k}\\ &=a_1\cdot\frac{1}{a_1}+a_2\left(\frac{1}{a_1}+\frac{1}{a_2}\right)+\ldots+a_n\left(\frac{1}{a_1}+\ldots+\frac{1}{a_n}\right)\\ &=\frac{1}{a_1}\left(a_1+\ldots+a_n\right)+\frac{1}{a_2}\left(a_2+\ldots+a_n\right)+\ldots+\frac{1}{a_n}\cdot a_n\\ &=\frac{1}{a_1}\sum_{k=1}^na_k+\frac{1}{a_2}\sum_{k=2}^na_k+\ldots+\frac{1}{a_n}\sum_{k=n}^na_k\\ &=\sum_{i=1}^n\left(\frac{1}{a_i}\cdot\sum_{k=i}^na_k\right). \end{align}$

Similarly, we have

$\displaystyle \begin{align} \sum_{i=1}^n\left(a_i\cdot\sum_{k=1}^i\frac{1}{a_k}\right)&=a_1\sum_{k=1}^n\frac{1}{a_k}+a_2\sum_{k=1}^{n-1}\frac{1}{a_k}+\ldots+a_n\sum_{k=1}^1\frac{1}{a_k}\\ &=a_1\cdot\left(\frac{1}{a_1}+\ldots+\frac{1}{a_n}\right)+a_2\left(\frac{1}{a_1}+\ldots+\frac{1}{a_{n-1}}\right)+\ldots+a_n\cdot\frac{1}{a_1}\\ &=\frac{1}{a_1}\left(a_1+\ldots+a_n\right)+\frac{1}{a_2}\left(a_1+\ldots+a_{n-1}\right)+\ldots+\frac{1}{a_n}\cdot a_1\\ &=\frac{1}{a_1}\sum_{k=1}^na_k+\frac{1}{a_2}\sum_{k=2}^{n-1}a_k+\ldots+\frac{1}{a_n}\sum_{k=1}^1a_k\\ &=\sum_{i=1}^n\left(\frac{1}{a_i}\cdot\sum_{k=i}^{n+1-i}a_k\right). \end{align}$

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