I proved this my trying to prove the contrapositive: if one of $x_n$ or $y_n$ is divergent, then $x_ny_n$must be also divergent. But we can have $x_n$ = n and $y_n=1/n$, where $x_ny_n=1$. Thus, a contradiction. Therefore, $x_ny_n$ being convergent does not imply $x_n,y_n$ to be both convergent.
Proof verification: If $x_ny_n$ is convergent, then $x_n,y_n$ are also convergent.
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real-analysis
proof-verification
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1You can also take $x_n = (-1)^n$ and $y_n = (-1)^{n+1}$ to get an even stronger conclusion. – 2017-01-18
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1How about $x_n=n$ and $y_n=0$? – 2017-01-18
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0Thanks for the comments! I see there are many counterexamples – 2017-01-18