We are familiar with the classic sum for Euler's constant γ:γ=limn→∞(∑k=1n1k−ln⁡(n))But, how is this one derived?:γ=limn→∞(12∑k=1∞(−1)k+1k(2k)!n2k−ln⁡(n))

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We are familiar with the classic sum for Euler&#039;s constant γ:γ=limn→∞(∑k=1n1k−ln⁡(n))But, how is this one derived?:γ=limn→∞(12∑k=1∞(−1)k+1k(2k)!n2k−ln⁡(n))

Sarai Dorsey · Accepted Answer

Consider sin2(n2):sin2(n2)=1−cos⁡(n)2=∑k=1∞(−1)k+12⋅(2k)!n2kNow, since n2k2k=∫0nt2k−1dt, we get:∑k=1∞(−1)k+1(2k)!n2k2k=∫0n1−cos⁡(t)tdtThe latter integral gives, by the definition of the cosine integral:∫0n1−cos⁡(t)tdt=γ+log⁡(n)−Ci(n)In order to get a sense of why Euler-Mascheroni constant γ appears here, consider another definition of cosine integral (in which it is manifest that limx→∞Ci(x)=0Ci(n)=−∫n∞cos⁡(t)tdtBoth definitions fulfill Ci′x=cos⁡(x)x. In order to establish that the integration constant is indeed the Euler-Mascheroni constant, we consider n=1:γ=∫011−cos⁡(t)tdt−∫1∞cos⁡(t)tdt=R(∫0∞[0&amp;lt;|t|&amp;lt;1]−eittdt)=γ

We are familiar with the classic sum for Euler's constant

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