If

\[A=\sum _{ n=1 }^{ \infty } \left( \sum _{ k=1 }^{ n^{ 2 } } \frac { d\left( k \right) -2M_{n}\left( k \right) }{ k } \right) n^{ -2 }\]

where \(d(k)\) is the number of divisors of \(k\) and \(M_{n}(k)\) is the number of divisors of \(k\) that are greater than \(n\), find \(\left\lfloor 100A \right\rfloor \).

Clue: Understand the derivation of \[{ \zeta (s) }^{ 2 }=\sum _{ k=1 }^{ \infty }{ \frac { d(k) }{ { k }^{ s } } } \]

I just realised that there was a typo in my previous question, so I'm reposting it. Sorry to those who wasted their time on a flawed problem.

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