Evaluation of $\sum_{n=1}^{\infty}\frac{1}{n(2n+1)}=2-2\ln(2)$

Question

I came across the following statements

$$\sum_{n=1}^{\infty} \frac{1}{n(2 n+1)}=2-2\ln 2 \qquad \tag{1}$$

$$\sum_{n=1}^{\infty} \frac{1}{n(3 n+1)}=3-\frac{3 \ln 3}{2}-\frac{\pi}{2 \sqrt{3}} \qquad \tag{2}$$

$$\sum_{n=1}^{\infty} \frac{1}{n(4 n+1)}=4-\frac{\pi}{2}-3 \ln 2 \qquad \tag{3}$$

$$\sum_{n=1}^{\infty} \frac{1}{n(6 n+1)}=6-\frac{\sqrt{3} \pi}{2}-\frac{3 \ln 3}{2}-2 \ln 2 \qquad \tag{4}$$

The (1) by partial fractions

$$\sum_{n=1}^{\infty}\frac{1}{n(2n+1)}=\sum_{n=1}^{\infty}\frac{1}{n}-\frac{2}{2n+1}$$

$$=\sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+\frac{1}{2}}$$

Recall the Digamma function

$$\psi(x+1)=\gamma+\sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+x}$$

Therefore

$$\sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+\frac{1}{2}}=\psi(1+\frac{1}{2})-\gamma$$

$$\sum_{n=1}^{\infty}\frac{1}{n(2n+1)}=\psi\left(\frac{3}{2}\right)-\gamma$$

In the same token we can derive the relation for the other three ralations. My Question is: can we calculate the values of the digamma function for those values without resorting in the Gauss´s Digamma formula?

$$\psi\left(\frac{r}{m}\right)=-\gamma-\ln (2 m)-\frac{\pi}{2} \cot \left(\frac{r \pi}{m}\right)+2 \sum_{n=1}^{\left\lfloor\frac{m-1}{2}\right\rfloor} \cos \left(\frac{2 \pi n r}{m}\right) \ln \sin \left(\frac{\pi n}{m}\right)$$

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I tried this approach also, but I think the resulting integral is divergent $$\sum_{n=1}^{\infty} \frac{1}{n(2 n+1)}=\sum_{n=1}^{\infty} \frac{1}{n}\int_{0}^{1}x^{2n}dx=\int_{0}^{1}\sum_{n=1}^{\infty} \frac{x^{2n}}{n}=-\int_{0}^{1}\ln(1-x^2)dx $$

Answer 1

$\def\o{\omega_k}$ $$ \sum_{n = 1}^\infty {\frac{1}{{n(2n + 1)}}} = 2\sum_{n = 1}^\infty\left(\frac1{2n}-\frac1{2n+1}\right)=2\left(1-\sum_{n = 1}^\infty\frac{(-1)^n}n\right)=2-2\log2. $$

Observe the additional factor $2$ comparing with a typo in your question.

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Inspired by this result we can elaborate the general solution:

We note that:

$$S_k=\sum_{n=1}^\infty\frac1{kn(kn+1)}=\sum_{n = 1}^\infty\left(\frac1{kn}-\frac1{kn+1}\right)=1-\frac1k \sum_{j=1}^k(1-\o^{-j})\log\left(1-\o^j\right),\tag1$$ where $\o=e^{\frac{2\pi i}k}$ is the $k$-th primitive root of unity.

To verify $(1)$ observe: $$ \frac1k\sum_{j=1}^k(\o^j)^p=\mathbb{1}_{p\equiv0\pmod k}. $$

Of course it is of interest to represent the solution (1) in terms of real parameters used in the question. This can be done as follows. Observe that we can omit in the sum the term $j=k$ and combine the complex conjugated terms $j$ and $k-j$ to obtain: $$ S_k=1-\frac1k\left\{\underbrace{\left[1+(-1)^k\right]}_{s_k}\log2+ \sum_{j=1}^{\left\lfloor\frac {k-1}2\right\rfloor} \left[(1-\o^{-j})\log\left(1-\o^j\right)+(1-\o^{j})\log\left(1-\o^{-j}\right)\right]\right\},\\ $$ where we singled out the unpaired term $2\log2$ for $j=\frac k2$ in the case of even $k$.

The sum can be rewritten as: $$\begin{align} &\sum_{j=1}^{\left\lfloor\frac {k-1}2\right\rfloor}\left\{\left(1-\cos\frac{2\pi j}k\right)\log\left(\left[1-e^{\frac{2\pi j}ki}\right]\left[1-e^{-\frac{2\pi j}ki}\right]\right)+i\sin\frac{2\pi j}k\log\left( \frac{1-e^{\frac{2\pi j}ki}}{1-e^{-\frac{2\pi j}ki}}\right) \right\}\\ &=\sum_{j=1}^{\left\lfloor\frac {k-1}2\right\rfloor}\left\{\left(1-\cos\frac{2\pi j}k\right)\log\left(2-2\cos\frac{2\pi j}k\right)+\pi\left(1-\frac{2j}k\right)\sin\frac{2\pi j}k \right\},\\ \end{align}$$ so that finally: $$\begin{align} S_k&=1-\frac1k\left\{s_k\log2+\sum_{j=1}^{\left\lfloor\frac {k-1}2\right\rfloor} \left[4\sin^2\frac{\pi j}k \log\left(2\sin\frac{\pi j}k\right)+ \pi\left(1-\frac{2j}k\right)\sin\frac{2\pi j}k\right]\right\}\\ &=1-\log2-\frac1k\left\{4\sum_{j=1}^{\left\lfloor\frac {k-1}2\right\rfloor}\sin^2\frac{\pi j}k \log\left(\sin\frac{\pi j}k\right)+ \frac\pi2\cot\frac\pi k\right\}.\tag2 \end{align}$$

The derivation of the last line is given below.

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Appendix

We are going to prove: $$Z_k=\sum_{0<j<\frac k2} \left(1-\frac{2j}k\right)\sin\frac{2\pi j}k=\frac12\cot\frac\pi k.$$

First we observe that due to symmetry: $$ Z_k=\frac12\sum_{0\le j<k}\left(1-\frac{2j}k\right)\sin\frac{2\pi j}k. $$

Next: $$\begin{align} 4Z_k\sin\frac\pi k&=\sum_{0\le j<k}\left(1-\frac{2j}k\right) \left(\cos\frac{\pi(2j-1)}k-\cos\frac{\pi(2j+1)}k\right)\\ &=\sum_{0\le j<k}\left(1-\frac{2j}k\right)\cos\frac{\pi(2j-1)}k -\sum_{1\le j\le k}\left(1+\frac2k-\frac{2j}k\right)\cos\frac{\pi(2j-1)}k\\ &=\underbrace{\cos\frac{\pi}k}_{j=0} +\underbrace{\cos\frac{\pi}k}_{j=k} -\frac2k\underbrace{\sum_{1\le j\le k}\cos\frac{\pi(2j-1)}k}_{=0}\\ &=2\cos\frac{\pi}k. \end{align}$$

The proof of the another identity used in $(2)$: $$\sum_{0<j<k}\sin^2\frac{\pi j}k=\frac k2$$ valid for $k\ge2$ is trivial.

Answer 2

One can observe that $$-\log(1-x^k)=\sum_{n\geq 1}\frac{x^{kn}}{n}$$ and integrating this with respect to $x$ on $[0,1]$ we get $$I_{k} =-\int_0^1\log(1-x^k)\,dx=\sum_{n\geq 1}\frac {1}{n(kn+1)}$$ Next we have $$I_k=-\int_{0}^{1}\log(1-x)\,dx-\int_0^1\log\frac{1-x^k}{1-x}\,dx=1-\int_0^1\log(1+x+\dots+x^{k-1})\,dx$$ and we can evaluate the integral for small values of $k$.

For general $k$ we need cyclotomic polynomial and their roots. Let $z_m=\exp(2m i\pi/k) $ for $m=1,2,\dots,k-1$ and then we have $$J_k=\int_0^1\log(1+x+\dots +x^{k-1})\,dx=\sum_{m=1}^{k-1}\int_0^1\log(x-z_m)\,dx$$ If $k$ is even then we have $z_{k/2}=-1$ and rest of the values of $z_m$ can be put into pairs of conjugates and we get $$J_k=\int_0^1\log(1+x)\,dx+\sum_{m<k/2}\int_0^1\log\left(x^2-2x\cos\frac{2m\pi}{k}+1\right)\,dx$$ which equals $$J_k=2\log 2 +1 - k+\sum_{m<k/2}\left\{4\sin^2\left(\frac{m\pi}{k}\right)\log\left(2\sin\left(\frac{m\pi}{k}\right)\right) +\frac{(k-2m)\pi}{k}\cdot\sin\left(\frac{2m\pi}{k}\right)\right\} $$

For odd $k$ all the roots $z_m$ get paired up with their conjugates and the expression for $J_k$ remains same as in case of even $k$ except for the term $2\log 2 $.

We thus have $$I_k=k-2\log 2-\sum_{m<k/2}\left\{4\sin^2\left(\frac{m\pi}{k}\right)\log\left(2\sin\left(\frac{m\pi}{k}\right)\right) +\frac{(k-2m)\pi}{k}\cdot\sin\left(\frac{2m\pi}{k}\right)\right\} $$ if $k$ is even and $$I_k=k- \sum_{m<k/2}\left\{4\sin^2\left(\frac{m\pi}{k}\right)\log\left(2\sin\left(\frac{m\pi}{k}\right)\right) +\frac{(k-2m)\pi}{k}\cdot\sin\left(\frac{2m\pi}{k}\right)\right\} $$ if $k$ is odd.

Using the above formula one can verify the given values in question. As others have noted you have a typo with the sum $I_2$ whose correct value is $2-2\log 2$.

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Here is an interesting fact which I noted. Since $I_k>0$ it follows that $$\int_0^1\log(1+x+\dots+x^{k-1})\,dx<1$$ The integrand is positive on $(0,1]$ and if $k$ is large it can take very large value $\log k$ and yet the integral is bounded for all $k$.

Answer 3

$$ \begin{align} \sum_{n=1}^\infty\frac1{n(2n+1)} &=2\lim_{N\to\infty}\sum_{n=1}^N\left(\frac1{2n}-\frac1{2n+1}\right)\tag1\\ &=2\lim_{N\to\infty}\sum_{n=1}^N\left(\frac1n-\left(\frac1{2n}+\frac1{2n+1}\right)\right)\tag2\\ &=2\lim_{N\to\infty}\left(\sum_{n=1}^N\frac1n-\sum_{n=2}^{2N+1}\frac1n\right)\tag3\\ &=2\lim_{N\to\infty}\left(1-\sum_{n=N+1}^{2N+1}\frac1n\right)\tag4\\ &=2\lim_{N\to\infty}\left(1-\sum_{n=N+1}^{2N+1}\frac Nn\frac1N\right)\tag5\\ &=2\left(1-\int_1^2\frac{\mathrm{d}x}x\right)\tag6\\[9pt] &=2(1-\log(2))\tag7 \end{align} $$ Explanation:
$(1)$: partial fractions
$\phantom{\text{(1):}}$ write infinite sum as a limit of partial sums
$(2)$: arithmetic
$(3)$: combine the inner parenthetical terms into one sum
$(4)$: cancel terms
$(5)$: multiply by $\frac NN$
$(6)$: convert Riemann Sum into integral
$(7)$: evaluate integral

Answer 4

It is very easy to use double integrals to handle this kinds of infinite sums. Let $$ f(x)=\sum_{n=1}^\infty\frac1{n(2n+1)}x^{2n+1} $$ and then $$ f'(x)=\sum_{n=1}^\infty\frac1{n}x^{2n}, f''(x)=\sum_{n=1}^\infty2x^{2n-1}=\frac{2x}{1-x^2}. $$ Then \begin{eqnarray} f(1)&=&\int_0^1f'(x)dx=\int_0^1\int_0^xf''(t)dtdx\\ &=&\int_0^1\int_t^1f''(t)dxdt=2\int_0^1\int_t^1\frac{t}{1-t^2}dxdt\\ &=&2\int_0^1\frac{t}{1-t^2}(1-t)dt=2\int_0^1\frac{t}{1+t}dt\\ &=&2(1-\ln2). \end{eqnarray}

Answer 5

The correct answer should be $2-2\log 2$. Note that $$ \sum\limits_{n = 1}^\infty {\frac{1}{{n(2n + 1)}}} = \mathop {\lim }\limits_{N \to + \infty } \left( {\sum\limits_{n = 1}^N {\frac{1}{n}} - 2\sum\limits_{n = 1}^N {\frac{1}{{2n + 1}}} } \right). $$ Now, in terms of the harmonic numbers, \begin{align*} \sum\limits_{n = 1}^N {\frac{1}{n}} - 2\sum\limits_{n = 1}^N {\frac{1}{{2n + 1}}} & = \sum\limits_{n = 1}^N {\frac{1}{n}} - 2\left( {-1+\sum\limits_{n = 1}^{2N + 1} {\frac{1}{n}} - \sum\limits_{n = 1}^N {\frac{1}{{2n}}} } \right) \\ & =2+ 2\sum\limits_{n = 1}^N {\frac{1}{n}} - 2\sum\limits_{n = 1}^{2N + 1} {\frac{1}{n}} =2+ 2H_N - 2H_{2N + 1} . \end{align*} To finish the proof, use the asymptotics $H_k = \log k + \gamma + o(1)$ as $k\to +\infty$.

Answer 6

Here is another method to add to the list. Consider the summation $$\sum\limits_{n=1}^{\infty }{\frac{1}{n\left( 2n+1 \right)}}$$ Note that $$\psi \left( -z \right)+\gamma \underset{z\to n}{\mathop{=}}\,\frac{1}{z-n}+{{H}_{n}}+\sum\limits_{k=1}^{\infty }{\left( {{\left( -1 \right)}^{k}}H_{n}^{k+1}-\zeta \left( k+1 \right) \right){{\left( z-n \right)}^{k}}}$$ for $n\ge 0$. So if we have a function $f$ that goes like $1/{{z}^{2}}$then taking $\left( \psi \left( -z \right)+\gamma \right)f\left( z \right)$over an infinitely large closed contour about $z=0$ just sums all the residues the total of which is zero. We have then $$\sum\limits_{n=1}^{\infty }{f\left( n \right)}=-\sum\limits_{f\left( z \right)}^{{}}{res\left( \psi \left( -z \right)+\gamma \right)f\left( z \right)}$$ where the sum on the right is over all residues due to $f$. In this case its very simple: there are two simple poles at $z=-1/2,0$. The residue at $z=0$ is easy (use the expansion above). We have $$\sum\limits_{n=1}^{\infty }{\frac{1}{n\left( 2n+1 \right)}}=2+\psi \left( \tfrac{1}{2} \right)+\gamma $$ At this point I’m just going to assume its well known that $\psi \left( \tfrac{1}{2} \right)=-\gamma -2\log \left( 2 \right)$