Intereting Posts

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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?:

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- The big $O$ versus little $o$ notation.
- Calculate: $ \lim_{x \to 0 } = x \cdot \sin(\frac{1}{x}) $

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

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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- Proving that $\left(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{n!}\right)$ has a limit

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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