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We are familiar with the classic sum for Euler's constant $\gamma$:

$$ \gamma=\lim_{n\to \infty}\left(\sum_{k=1}^{n}\frac{1}{k}-\ln(n)\right) .$$

But, how is this one derived?:

$$ \gamma=\lim_{n\to \infty}\left(\frac12\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k(2k)!}n^{2k}-\ln(n)\right) .$$

I thought perhaps it may have something to do with the Bernoulli numbers or even Zeta because there are terms which appear in their identities (such as the formula for $\zeta(2n)$), but I am not sure.

Thanks for any input.

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$$\frac12\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k(2k)!}n^{2k}=\gamma-\mathrm{Ci}(n)+\ln\,n$$, where $\mathrm{Ci}(z)$ is the cosine integral. –  J. M. Dec 22 '11 at 12:37
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An equivalent expression for your limit is $$\lim_{z\to \infty}\left(\int_0^z \frac{1-\cos\,t}{t}\mathrm dt-\ln\,z\right)=\gamma$$ –  J. M. Dec 22 '11 at 12:40
    
Yet another way to express your limit: $$\int_0^1 \frac{1-\cos\,t}{t}\mathrm dt-\int_1^\infty \frac{\cos\,t}{t}\mathrm dt=\gamma$$ –  J. M. Dec 22 '11 at 12:53
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Thank you very much. I was unfamiliar with that series. –  Cody Dec 22 '11 at 12:54
    
@J.M. Why not post this as an answer? You'll get some upvote loves... –  Srivatsan Dec 22 '11 at 21:49
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up vote 2 down vote accepted

Consider $\sin^2\left(\frac{n}{2} \right)$: $$ \sin^2\left(\frac{n}{2} \right) = \frac{1-\cos(n)}{2} = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{2 \cdot (2k)!} n^{2k} $$ Now, since $\frac{n^{2k}}{2 k} = \int\limits_0^n t^{2k-1} \mathrm{d} t$, we get: $$ \sum_{k=1}^\infty \frac{(-1)^{k+1}}{(2k)!} \frac{n^{2k}}{2 k} = \int_0^n \frac{1-\cos(t)}{t} \mathrm{d} t $$ The latter integral gives, by the definition of the cosine integral: $$ \int_0^n \frac{1-\cos(t)}{t} \mathrm{d} t = \gamma + \log(n) - \operatorname{Ci}(n) $$

In order to get a sense of why Euler-Mascheroni constant $\gamma$ appears here, consider another definition of cosine integral (in which it is manifest that $\lim_{x \to \infty} \operatorname{Ci}(x) = 0$): $$ \operatorname{Ci}(n) = -\int_{n}^\infty \frac{\cos(t)}{t} \mathrm{d} t $$

Both definitions fulfill $\operatorname{Ci}^\prime(x) = \frac{\cos(x)}{x}$. In order to establish that the integration constant is indeed the Euler-Mascheroni constant, we consider $n=1$: $$ \begin{eqnarray} \gamma &=& \int_0^1 \frac{1 - \cos(t)}{t} \mathrm{d} t - \int_1^\infty \frac{\cos(t)}{t} \mathrm{d} t = \Re\left( \int_0^\infty \frac{[0<|t|<1]-\mathrm{e}^{i t}}{t} \mathrm{d} t \right) \\ &\stackrel{t \to i t}{=}& \Re\left( \int_0^\infty \frac{[0<|t|<1]-\mathrm{e}^{-t}}{t} \mathrm{d} t \right) \stackrel{\text{by parts}}{=} \Re\left( -\int_0^\infty \mathrm{e}^{-t} \ln(t) \mathrm{d} t \right) = \gamma \end{eqnarray} $$ The last equality follows from the definition of the constant.

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