How do I prove this identity?∑n=1∞1ns∑n=1∞μ(n)ns=1

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llevochalecoiozq · Accepted Answer

The idea needed here is Dirichlet convolution.The convolution identity states that if(∑j=1∞ajjs)(∑k=1∞bkks)=∑ell=1∞cllsthen∑k,∣,nakbnk=cnwhere the sum ranges over all positive divisors of nMore specifically, your original identity is equivalent to the convolution identity∑k,∣,nμ(k)=[n=1]where the two functions being convolved are the constant function 1 and the Möbius function μ(k) ([p] here is the Iverson bracket; it evaluates to 1 if p is true, and 0 if p is false.)For n=1, the identity is easily verified. To see what happens when n&amp;gt;1, take the &#039; factorizationn=∏k=1rpkmkIt is known that the summatory function of a multiplicative function is itself multiplicative; thus, letting ξ(n)=∑k,∣,nμ(k), we have ξ(qr)=ξ(q)ξ(r). We can thus restrict to the caseξ(pm)=μ(1)+μ(p)+⋯+μ(pm)=1−1+0+⋯+0=0and since ξ(n)=ξ(p1m1)ξ(p2m2)⋯ξ(prmr) for n&amp;gt;1

How do I prove this identity? \(\displaystyle{\sum_{{{n}={1}}}^{{\infty}}}{\frac{{{1}}}{{{n}^{{s}}}}}{\sum_{{{n}={1}}}^{{\infty}}}{\frac{{\mu{\left({n}\right)}}}{{{n}^{{s}}}}}={1}\)

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