Finding $\sum_{k=1}^{\infty} \left[\frac{1}{2k}-\log \left(1+\frac{1}{2k}\right)\right]$

Question

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

I know that $\displaystyle\sum_{k=1}^{\infty} \left[\frac{1}{k} -\log\left(1+\dfrac{1}{k}\right)\right]=\gamma$, where $\gamma$ is the Euler–Mascheroni constant.

But I could not manipulate this series.

Answer 1

Start with the infinite product expansion of Gamma function,

$$\frac{1}{\Gamma(z)} = z e^{\gamma z}\prod_{k=1}^\infty \left(1 + \frac{z}{k}\right) e^{-\frac{z}{k}}$$

Taking logarithm and rearrange terms, one get

$$\begin{align} & -\log\Gamma(z) = \log z + \gamma z + \sum_{k=1}^\infty \left[\log\left(1+\frac{z}{k}\right) - \frac{z}{k}\right]\\ \implies & \sum_{k=1}^\infty \left[\frac{z}{k} - \log\left(1+\frac{z}{k}\right) \right] = \log z + \gamma z + \log\Gamma(z) \end{align}$$ Taking $z = \frac12$ and use the known value of $\Gamma\left(\frac12\right) = \sqrt{\pi}$, we get

$$\sum_{k=1}^\infty \left[\frac{1}{2k} - \log\left(1+\frac{1}{2k}\right) \right] = \frac12\left(\gamma + \log\pi\right) - \log 2 $$

Answer 2

Consider the series \begin{align} S = \sum_{k=1}^{\infty} \left[ \frac{1}{2k} - \ln\left(1 + \frac{1}{2k}\right) \right] \end{align} for which, by using the logarithm in series form, it becomes \begin{align} S &= \sum_{k=1}^{\infty} \left[ \frac{1}{2k} - \frac{1}{2k} + \sum_{n=2}^{\infty} \frac{(-1)^{n}}{n \, (2k)^{n}} \right] \\ &= \sum_{k=1}^{\infty} \sum_{n=2}^{\infty} \frac{(-1/2)^{n}}{n \, k^{n}} \\ &= \sum_{n=2}^{\infty} \frac{(-1/2)^{n}}{n} \, \zeta(n) \\ &= \sum_{n=1}^{\infty} \frac{(-1/2)^{n+1}}{n+1} \, \zeta(n+1). \end{align} Now the generating function for the zeta function is given by \begin{align} \sum_{n=1}^{\infty} \zeta(n+1) (-1)^{n+1} x^{n} = \gamma + \psi(x+1) \end{align} and by integration leads to \begin{align} \sum_{n=1}^{\infty} \frac{(-x)^{n+1}}{n+1} \, \zeta(n+1) = \gamma x + \ln\Gamma(x+1). \end{align} Letting $x = 1/2$ in this series leads to the value of the series for $S$, namely \begin{align} \sum_{k=1}^{\infty} \left[ \frac{1}{2k} - \ln\left(1 + \frac{1}{2k}\right) \right] = \frac{1}{2} \ln\left(\frac{\pi}{4}\right) + \frac{\gamma}{2}. \end{align}

Answer 3

If I have well understood, you want the limit of $\displaystyle S_n=\sum_{k=1}^n (\frac{1}{2k}-\log(1+\frac{1}{2k}))$.

We have $$\sum_{k=1}^n \frac{1}{k}=\log(n+1)+\gamma+\varepsilon_n$$ with $\varepsilon_n\to 0$ as $n\to +\infty$. Now: $$S_n=\frac{1}{2}(\sum_{k=1}^n\frac{1}{k})-\sum_{k=1}^n \log(2k+1)+\sum_{k=1}^n \log(2k)=\frac{1}{2}(\sum_{k=1}^n\frac{1}{k})-\sum_{k=1}^n (\log(2k+1)+\log(2k))+2\sum_{k=1}^n \log(2k)$$ Hence: $$S_n=\frac{\gamma}{2}+\frac{1}{2}\log(n+1)+\frac{\varepsilon_n}{2}-\log(2n+1)!+2n\log 2+2\log n!$$

Thus: $$S_n=\frac{\gamma}{2}+\frac{\varepsilon_n}{2}+\log A_n$$ With $\displaystyle A_n=\frac{2^{2n}(n!)^2\sqrt{n+1}}{(2n+1)!}$. Now use Stirling's Formula to finish.

Answer 4

Observe that, by absolute convergence: $$\begin{align} \sum_{k=1}^{\infty} \left(\frac{1}{2k} -\log\left(1+\dfrac{1}{2k}\right)\right)& =\sum_{k=1}^{\infty} \dfrac{1+(-1)^k}{2}\left(\frac{1}{k} -\log\left(1+\dfrac{1}{k}\right)\right)\\\\ &=\frac{1}{2}\sum_{k=1}^{\infty} \!\left(\frac{1}{k} -\log\left(1+\dfrac{1}{k}\!\right)\right)\!+\!\frac{1}{2}\!\sum_{k=1}^{\infty} (-1)^{k-1}\left(\!\frac{1}{k} -\log\!\left(1+\dfrac{1}{k}\!\right)\!\right) \\\\ &=\frac{\gamma}{2}+\!\frac{1}{2}\!\sum_{k=1}^{\infty} (-1)^{k-1}\left(\!\frac{1}{k} -\log\!\left(1+\dfrac{1}{k}\!\right)\!\right). \end{align} $$ Then, as $N$ is great, write $$ \begin{align} \sum_{k=1}^{N} (-1)^{k-1}\left(\!\frac{1}{k} -\log\!\left(1+\dfrac{1}{k}\!\right)\!\right)& =\sum_{k=1}^{N}(-1)^{k-1} \frac{1}{k}-\sum_{k=1}^{N} (-1)^{k-1}\log\!\left(1+\dfrac{1}{k}\!\right)\\\\ &=\sum_{k=1}^{N}(-1)^{k-1} \frac{1}{k}-\log \left(\prod_{k=1}^{N} \left(1+\dfrac{1}{k}\!\right)^{(-1)^{k-1}}\right) \tag2 \end{align} $$ giving $$ \begin{align} \sum_{k=1}^{\infty} (-1)^{k-1}\left(\!\frac{1}{k} -\log\!\left(1+\dfrac{1}{k}\!\right)\!\right)=\log 2-\log \left(\frac{\pi}{2}\right) \end{align} $$ where we have used Wallis' product formula for $\pi$.

Finally we obtain

$$ \sum_{k=1}^{\infty} \left(\frac{1}{2k} -\log\left(1+\dfrac{1}{2k}\right)\right)=\frac{\gamma}{2}-\frac{1}{2}\log \left(\frac{4}{\pi}\right). $$

Answer 5

You can do it by calulating the n-th partial sum and taking the limit:

Let $S_n=\sum_{k=1}^{n} \left[\frac{1}{2k} -\log\left(1+\dfrac{1}{2k}\right)\right]$. Then:

$$ S_n=\frac{1}{2}H_n -\log\left(\prod_{k=1}^n\dfrac{2k+1}{2k}\right)=\frac{1}{2}H_n -\log\left(\dfrac{(2n+1)!!}{(2n)!!}\right)=\frac{1}{2}H_n -\log\left(\dfrac{(2n+1)!}{2^{2n}(n!)^2}\right) $$

By Stirlings approximation, we have

$$ n!=\sqrt{2\pi}\cdot n^{n+\frac{1}{2}}\cdot e^{-n}\cdot e^{\frac{\theta_n}{12n}} $$

With $0\lt\theta_n\lt1$. Making this substitution in the fomula above, we get:

$$ S_n=\frac{1}{2}H_n -\log\left(\dfrac{\sqrt{2\pi}\cdot(2n+1)^{2n+1+\frac{1}{2}}\cdot e^{-2n-1}\cdot e^{\frac{\theta_{2n+1}}{12(2n+1)}}}{2^{2n}\cdot 2\pi\cdot n^{2n+1} \cdot e^{-2n}\cdot e^{\frac{\theta_n}{6n}}}\right)=\frac{1}{2}H_n -\log\left(\dfrac{2\cdot(1+\frac{1}{2n})^{2n+1}\cdot \sqrt{n+\frac{1}{2}}\cdot e^{\frac{\theta_{2n+1}}{12(2n+1)}}}{\sqrt{\pi}\cdot e\cdot e^{\frac{\theta_n}{6n}}}\right)=\frac{1}{2}H_n -\frac{1}{2}\log\left(n+\frac{1}{2}\right)-\log\left(\dfrac{2\cdot(1+\frac{1}{2n})^{2n+1}\cdot e^{\frac{\theta_{2n+1}}{12(2n+1)}}}{\sqrt{\pi}\cdot e\cdot e^{\frac{\theta_n}{6n}}}\right) $$

So by taking the limit, we get:

$$ S=\lim_{n \to \infty}S_n=\lim_{n \to \infty}{\left[\frac{1}{2}H_n -\frac{1}{2}\log\left(n+\frac{1}{2}\right)-\log\left(\dfrac{2\cdot(1+\frac{1}{2n})^{2n+1}\cdot e^{\frac{\theta_{2n+1}}{12(2n+1)}}}{\sqrt{\pi}\cdot e\cdot e^{\frac{\theta_n}{6n}}}\right)\right]=\frac{\gamma}{2}-\log(2)+\frac{1}{2}\log{(\pi)}} $$

Which seems to be true since Wolfram alpha gives:

$$ \sum_{k=1}^{\infty} \left[\frac{1}{2k} -\log\left(1+\dfrac{1}{2k}\right)\right] \approx 0.1678255948155212079577375992595540032692269400673623 $$

While:

$$ \frac{\gamma}{2}-\log(2)+\frac{1}{2}\log{(\pi)} \approx 0.1678255948155212079577375992595540032692269400673623 $$

Answer 6

Notice that $$S=\sum^{\infty}_{k=1}\Big(\frac{1}{2k}-\frac{1}{2}\log(1+\frac{1}{k})+\frac{1}{2}\log(1+\frac{1}{k})-\frac{2}{2}\log(1+\frac{1}{2k})\Big)$$ Therefore $$S=\frac{1}{2}\gamma+\frac{1}{2}\sum^{\infty}_{k=1}\log\Big(\frac{1+\frac{1}{k}}{(1+\frac{1}{2k})^2}\Big)$$ In other words \begin{align} S=\frac{1}{2}\gamma+\frac{1}{2}\log\Big(\prod^{\infty}_{k=1}\frac{1+\frac{1}{k}}{(1+\frac{1}{2k})^2}\Big)&=\frac{1}{2}\gamma+\frac{1}{2}\log\Big(\lim_{N\to\infty}\prod^{N}_{k=1}\frac{1+\frac{1}{k}}{(1+\frac{1}{2k})^2}\Big)\\&=\frac{1}{2}\gamma+\frac{1}{2}\log\Big(\lim_{N\to\infty}\prod^{N}_{k=1}\frac{k+1}{k}\frac{(2k)^2}{(2k+1)^2}\Big)\\ &=\frac{1}{2}\gamma+\frac{1}{2}\log\Big(\lim_{N\to\infty}\frac{\pi\Gamma(N+1)\Gamma(N+2)}{4\Gamma^2(N+3/2)}\Big)\\ &=\frac{1}{2}\gamma+\frac{1}{2}\log(\frac{\pi}{4}) \end{align} We have used the limiting value $$\lim_{N\to\infty}\frac{\Gamma(N+1)\Gamma(N+2)}{\Gamma^2(N+3/2)}=1$$