Evaluate the expression S=∑n=1∞log2(1+1n)⋅logn+1(2)S = \sum_{n=1}^{\infty} \log_2 \left( 1 + \frac{1}{n} \right) \cdot \log_{n+1} (2)S=∑n=1∞log2(1+n1)⋅logn+1(2).
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Converges to infinity