prove-that-log-2-2-2-n-1-n-n-1-2-2-n-log-2-n-1-n-n-1-3-2-n-n-1-n-n-1-4-2-n-23-8-6-2-2-3-1-18-log-6-2-m-A-

Question Number 155136 by amin96 last updated on 25/Sep/21

p r o v e t h a t

\frac{\log^{2} (2)}{2} \underset{n = 1}{\sum^{\infty}} \frac{φ (n)}{{(n + 1)}^{2} 2^{n}} + \log (2) \underset{n = 1}{\sum^{\infty}} \frac{φ (n)}{{(n + 1)}^{3} 2^{n}} + \underset{n = 1}{\sum^{\infty}} \frac{φ (n)}{{(n + 1)}^{4} 2^{n}} =

= \frac{23}{8} ζ (6) - 2 ζ^{2} (3) - \frac{1}{18} \log^{6} (2)

m . A