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$\begingroup$

$$A=\left[\frac{1}{i+j}\right]=\left(\begin{matrix}\frac{1}{1+1}&\frac{1}{1+2}&\cdots&\frac{1}{1+n}&\\\frac{1}{2+1}&\frac{1}{2+2}&\cdots&\frac{1}{2+n}\\\vdots&\vdots&\ddots&\vdots\\\frac{1}{n+1}&\frac{1}{n+2}&\cdots&\frac{1}{n+n}\end{matrix}\right)$$

My textbook exercise says yes, but I can't prove it.

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Yes, this is similar to a Hilbert matrix, and a clever trick does it: Note that the $(j,k)$ element is $a_{jk}=\int_0^1 x^{j+k-1}\,dx$, and so $$\sum_{j,k} a_{jk}\bar u_ju_k=\int_0^1\Bigl|\sum_j u_jx^j\Bigr|^2x^{-1}\,dx\ge0.$$

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    Please check this proof http://math.stackexchange.com/questions/1515638/sum-limits-j-k-a-jk-bar-u-ju-k-mathop-limits-int-01-sum/1515997?noredirect=1#comment3094802_1515997 By this proof, your proof is'nt true.2015-11-08
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    @H.S Right, but that was just a simple typo. Fixed now.2015-11-09
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    check this...http://math.stackexchange.com/questions/1515954/can-we-prove-that-all-eigenvalues-of-a-are-positive/1516004#15160042015-11-09
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    @Harald Hanche-Olsen - Thanks.2015-11-09
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    @soumitra Yes, that is another, perfectly valid, way of doing it.2015-11-09