How do we findS=∑k=1∞[12k−log⁡(1+12k)]

Question

Donald Erickson · Accepted Answer

Observe that, by absolute convergence:  ∑k=1∞(12k−log⁡(1+12k))=∑k=1∞1+(−1)k2(1k−log⁡(1+1k))  =12∑k=1∞(1k−log⁡(1+1k))+12∑k=1∞(−1)k−1(1k−log⁡(1+1k))  =γ2+12∑k=1∞(−1)k−1(1k−log⁡(1+1k))  Then, as N is great, write  ∑k=1N(−1)k−1(1k−log⁡(1+1k))=∑k=1N(−1)k−11k−∑k=1N(−1)k−1log⁡(1+1k)  =∑k=1N(−1)k−11k−log⁡(∏k=1N(1+1k)(−1)k−1)  giving  ∑k=1∞(−1)k−1(1k−log⁡(1+1k))=log⁡2−log⁡(π2)  Finally we obtain

Ulgelmorgs · Accepted Answer

Consider the series  S=∑k=1∞[12k−ln⁡(1+12k)]  for which, by using the logarithm in series form, it becomes  S=∑k=1∞[12k−12k+∑n=2∞(−1)nn(2k)n]  =∑k=1∞∑n=2∞(−12)nnkn  =∑n=2∞(−12)nnζ(n)  =∑n=1∞(−12)n+1n+1ζ(n+1)  Now the generating function for the zeta function is given by  ∑n=1∞ζ(n+1)(−1)n+1xn=γ+ψ(x+1)  and by integration leads to  ∑n=1∞(−x)n+1n+1ζ(n+1)=γx+ln⁡Γ(x+1)  Letting x=12 in this series leads to the value of the series for S, namely  ∑k=1∞[12k−ln⁡(1+12k)]=12ln⁡(π4)+γ2

How do we find S=\sum_{k=1}^\infty[\frac{1}{2k}-\log(1+\frac{1}{2k})]

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