sum[(1/2)^(1/n)/n^2,{n,1,infty}],sum[zeta(m+2) (ln(1/2))^m/m!,{m,0,infty}]sum[(1/2)^(1/n)/n^2,{n,1,infty}],sum[zeta(m+2) (ln(1/2))^m/m!,{m,0,infty}]

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