Evaluate ∑n=2∞log2(1−1n2)

Let S=∑n=2∞log2(1−1n2)=∑n=2∞(log2(1−1n)+log2(1+1n))=∑n=2∞(log2(n−1n)−log2(nn+1))Let Vr=log2(r−1r)⇒Vr+1=log2(rr+1)⇒S=∑r=2∞(Vr−Vr+1)=V2=log2(12)=−1

Evaluate ∑n=2∞log2(1−1n2)

Let S=∑n=2∞log2(1−1n2)=∑n=2∞(log2(1−1n)+log2(1+1n))=∑n=2∞(log2(n−1n)−log2(nn+1))Let Vr=log2(r−1r)⇒Vr+1=log2(rr+1)⇒S=∑r=2∞(Vr−Vr+1)=V2=log2(12)=−1

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Evaluate $\sum_{n = 2}^{\infty} \log_{2} (1 - \frac{1}{n^{2}})$

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Detailed Solution

Let $S = \sum_{n = 2}^{\infty} \log_{2} (1 - \frac{1}{n^{2}}) = \sum_{n = 2}^{\infty} (\log_{2} (1 - \frac{1}{n}) + \log_{2} (1 + \frac{1}{n})) = \sum_{n = 2}^{\infty} (\log_{2} (\frac{n - 1}{n}) - \log_{2} (\frac{n}{n + 1}))$
Let $V_{r} = \log_{2} (\frac{r - 1}{r}) \Rightarrow V_{r + 1} = \log_{2} (\frac{r}{r + 1})$
$\Rightarrow S = \sum_{r = 2}^{\infty} (V_{r} - V_{r + 1}) = V_{2} = \log_{2} (\frac{1}{2}) = - 1$