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Prove that the given infinite series has the stated value

Prove that the following sum converges and has the given value.

    \[ \sum_{n=2}^{\infty} \frac{\log \left( \left( 1 + \frac{1}{n} \right)^n (1+n) \right)}{(\log (n^n))(\log (n+1)^{n+1})} = \log_2 \sqrt{e}. \]


Again, we simplify and apply the theorem on the sums of telescoping series (Theorem 10.7 on page 386 of Apostol),

    \begin{align*}  \sum_{n=2}^{\infty} \frac{\log \left( 1 + \frac{1}{n} \right)^n (1+n)}{(\log (n^n))(\log (n+1)^{n+1})} &= \sum_{n=2}^{\infty} \frac{n \log \left( 1 + \frac{1}{n} \right) + \log(1+n)}{n \log n \cdot (n+1) \log (n+1)} \\[9pt]  &= \sum_{n=2}^{\infty} \frac{n \log(n+1) - n \log n + \log (n+1)}{n \log n \cdot (n+1) \log (n+1)} \\[9pt]  &= \sum_{n=2}^{\infty} \frac{(n+1)\log(n+1) - n \log n}{n \log n \cdot (n+1) \log (n+1)} \\[9pt]  &= \sum_{n=2}^{\infty} \left( \frac{1}{n \log n} - \frac{1}{(n+1)\log (n+1)} \right) \\[9pt]  &= \sum_{n=1}^{\infty} \left( \frac{1}{(n+1)\log(n+1)} - \frac{1}{(n+2)\log(n+2)} \right) \\[9pt]  &= b_1 - L  \\[9pt]   &= \frac{1}{2 \log 2} \\[9pt]  &= \frac{1}{2} \left( \frac{\log e}{\log 2} \right) \\[9pt]  &= \frac{1}{2} \log_2 e \\[9pt]  &= \log_2 \sqrt{e}. \end{align*}

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