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

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 math.stackexchange.com/questions/1515638/… By this proof, your proof is'nt true. – H.S Nov 8 '15 at 21:19
1  
@H.S Right, but that was just a simple typo. Fixed now. – Harald Hanche-Olsen Nov 9 '15 at 16:42
    
check this...math.stackexchange.com/questions/1515954/… – Black -horse Nov 9 '15 at 16:47
    
@Harald Hanche-Olsen - Thanks. – H.S Nov 9 '15 at 17:06
    
@soumitra Yes, that is another, perfectly valid, way of doing it. – Harald Hanche-Olsen Nov 9 '15 at 17:09

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