# How to show that $\sum_{n=1}^{\infty} \frac{1}{(2n+1)(2n+2)(2n+3)}=\ln(2)-1/2$?

How i can prove that

$$\sum_{n=1}^{\infty} \frac{1}{(2n+1)(2n+2)(2n+3)}=\ln(2)-1/2$$

And $$\sum_{n=1}^{\infty} \frac{1}{(4n+1)(4n+2)(4n+3)(4n+4)}=\frac{1}{4}\left(\ln(2) - \frac{\pi}{6}\right).$$

Thanks in advance.

• Welcome to MSE! Have you tried anything and can share your thoughts and attempts? Regards – Amzoti Apr 20 '13 at 23:52
• The first sum is $\ln\left(2\right) - 2/3$. When the $n$-index start at $0$, the result is $\ln\left(2\right) - 1/2$. See my answer. I didn't check the second one. – Felix Marin Aug 26 '13 at 9:46

## 5 Answers

An alternative approach; compute

$$f(x)=\sum_{n=0}^\infty \frac{x^{2n+3}}{(2n+3)(2n+2)(2n+1)}$$

This has the property that $f'''(x)= \sum_{n=0}^\infty x^{2n} = \frac{1}{1-x^2}$. Integrate once (partial fractions) to get logarithms. Integrate twice more (by parts) and evaluate at 1 to get the answer $f(1)$.

• +1. This is a nice and probably an easier approach than mine, though the reason why both work are similar if not the same. – user17762 Apr 21 '13 at 0:15
• Absolutely! I started writing "ultimately equivalent to the other answer" but then thought this would merit extra discussion so I left it out, being lazy... – Sharkos Apr 21 '13 at 0:17

I assume both your summation starts from $0$ instead of $1$. We have $$a_n = \dfrac1{(2n+1)(2n+2)(2n+3)} = \dfrac1{2(2n+1)} + \dfrac1{2(2n+3)} - \dfrac1{2n+2}$$ This gives us $$a_n = \dfrac12 \int_0^1x^{2n} dx + \dfrac12 \int_0^1x^{2n+2} dx - \int_0^1 x^{2n+1} dx = \dfrac12 \int_0^1 x^{2n}(1-x)^2dx$$ Hence, $$\sum_{n=0}^{\infty} a_n = \dfrac12 \int_0^1 \dfrac{(1-x)^2}{1-x^2}dx = \dfrac12 \int_0^1 \dfrac{1-x}{1+x}dx = \int_0^1 \dfrac{dx}{1+x} - \dfrac12 \int_0^1dx = \log(2) - \dfrac12$$ The same idea works for the other series as well and I will leave it to you to work out the details. Be careful on two counts, while following the above technique:

$1$. Note that I wrote $\dfrac1{2n+2}$ as $\displaystyle \int_0^1 x^{2n+1} dx$ and not as $\displaystyle \dfrac12 \int_0^1 x^n dx$. Though both are valid ways to obtain $\dfrac1{2n+2}$, if you do the second way, when you sum it up, you are changing the order of summation and hence will get a different incorrect answer.

$2$. Also, make sure to justify the change of integration and limits.

This is similar, but without partial fractions!

$$(1)$$

\begin{aligned} \sum_{n\geq 0}\frac{1}{(2n+1)(2n+2)(2n+3)} & =\sum_{n\geq 0} \frac{\Gamma(2n+1)}{\Gamma(2n+4)} \\& =\sum_{n\geq 0}\frac{1}{2}\mathrm{B}(2n+1, \,3) \\& = \frac{1}{2}\sum_{n\geq 0}\int_{0}^{1}x^{2n}(1-x)^2\; dx\\& = \frac{1}{2}\int_{0}^{1} \frac{1-x}{1+x} \;{dx} \\& = \ln 2-\frac{1}{2} \end{aligned}

$$(2)$$

\begin{aligned} \sum_{n\geq 0}\frac{1}{(4n+1)(4n+2)(4n+3)(4n+4)} & =\sum_{n\geq 0} \frac{\Gamma(4n+1)}{\Gamma(4n+5)} \\& =\sum_{n\geq 0}\frac{1}{6}\mathrm{B}(4n+1, \,4) \\& = \frac{1}{6}\sum_{n\geq 0}\int_{0}^{1}x^{4n}(1-x)^3\; dx\\& = \frac{1}{6}\int_{0}^{1} \frac{(1-x)^2}{(1+x^2)(1+x)} \; dx \\& = \frac{\ln 2}{4}-\frac{\pi}{24} \end{aligned}

• +1. This is neat. Slightly circumvents the possible mistake of rearranging sums, that could happen in my method. – user17762 Apr 21 '13 at 1:34
• @user17762 Thank you! – L. F. Apr 21 '13 at 1:37

\begin{align} &\sum_{n=0}^\infty\frac1{(2n+1)(2n+2)(2n+3)}\\ &=\sum_{n=0}^\infty\left(\frac{1/2}{2n+1}+\frac{-1}{2n+2}+\frac{1/2}{2n+3}\right)\\ &=-\frac{\frac12}{1}+\frac{\frac12}{1}\tag{add 0}\\ &\hphantom{=-\frac{\frac12}{1}}+\frac{\frac12}{1}-\frac12+\frac{\frac12}{3}\tag{n=0}\\ &\hphantom{=-\frac{\frac12}{1}+\frac{\frac12}{1}-\frac12}+\frac{\frac12}{3}-\frac14+\frac{\frac12}{5}\tag{n=1}\\ &\hphantom{=-\frac{\frac12}{1}+\frac{\frac12}{1}-\frac12+\frac{\frac12}{3}-\frac14}+\frac{\frac12}{5}-\frac16+\dots\tag{n=2}\\ &=\color{#C00000}{-\frac12}+\color{#00A000}{1-\frac12\,+\frac13-\frac14+\frac15-\frac16+\dots}\\ &=\color{#00A000}{\log(2)}\color{#C00000}{-\frac12} \end{align}

\begin{align} &\sum_{n=0}^\infty\frac1{(4n+1)(4n+2)(4n+3)(4n+4)}\\ &=\sum_{n=0}^\infty\left(\frac{1/6}{4n+1}+\frac{-1/2}{4n+2}+\frac{1/2}{4n+3}+\frac{-1/6}{4n+4}\right)\\ &=\sum_{n=0}^\infty\left(\frac{-1/6}{4n+1}+\frac{1/6}{4n+3}\right)+\left(\frac{1/3}{4n+1}+\frac{1/3}{4n+3}\right)\\ &\hphantom{\sum_{n=0}^\infty}+\left(\frac{-1/6}{4n+2}+\frac{1/6}{4n+4}\right)+\left(\frac{-1/3}{4n+2}+\frac{-1/3}{4n+4}\right)\\ &=\color{#00A000}{\sum_{n=0}^\infty\left(\frac{-1/6}{4n+1}+\frac{1/6}{4n+3}\right)}+\color{#0000FF}{\sum_{n=0}^\infty\left(\frac{-1/6}{4n+2}+\frac{1/6}{4n+4}\right)}\\ &+\color{#C00000}{\sum_{n=0}^\infty\left(\frac{1/3}{4n+1}+\frac{1/3}{4n+3}\right)+\left(\frac{-1/3}{4n+2}+\frac{-1/3}{4n+4}\right)}\\ &=\color{#00A000}{-\frac\pi{24}}\color{#0000FF}{-\frac{\log(2)}{12}}\\ &\color{#C00000}{+\frac{\log(2)}{3}}\\ &=\frac{\log(2)}{4}-\frac\pi{24} \end{align} $\color{#00A000}{\text{Leibniz Series}}$, $\color{#0000FF}{\text{Alternating Harmonic Series}}$, and $\color{#C00000}{\text{Alternating Harmonic Series}}$.

• I am not completely sure but looks like you are rearranging terms for the second summation (?) – user17762 Apr 21 '13 at 1:47
• @user17762: sure, but only splitting a block of four terms into a block of eight term, then reorganizing the block of eight. All of it perfectly fine. – robjohn Apr 21 '13 at 3:34
• In the following line $$\sum_{n=0}^\infty\left(\frac{-1/6}{4n+1}+\frac{1/6}{4n+3}\right)+\left(\frac{1/3}{4n+1}+\frac{1/3}{4n+3}\right)\\ +\sum_{n=0}^\infty\left(\frac{-1/6}{4n+2}+\frac{1/6}{4n+4}\right)+\left(\frac{-1/3}{4n+2}+\frac{-1/3}{4n+4}\right)$$ both the individual sums are $\infty$ (?). – user17762 Apr 21 '13 at 3:37
• @user17762: I was too exuberant with my sigmas. It should have been one sum broken over two lines. – robjohn Apr 21 '13 at 9:39

\begin{align} &\sum_{n = 0}^{\infty} {1 \over \left(2n + 1\right)\left(2n + 2\right)\left(2n + 3\right)} = {1 \over 4}\sum_{n = 0}^{\infty} \left({1 \over n + 1/2} - {2 \over n + 1} + {1 \over n + 3/2}\right) \\[3mm]&= {1 \over 8}\left\lbrack \sum_{n = 0}^{\infty}{1 \over \left(n + 1/2\right)\left(n + 1\right)} - \sum_{n = 0}^{\infty}{1 \over \left(n + 1\right)\left(n + 3/2\right)} \right\rbrack \\[3mm]&= {1 \over 8}\left\lbrack {\Psi\left(1/2\right) - \Psi\left(1\right) \over 1/2 - 1} - {\Psi\left(1\right) - \Psi\left(3/2\right) \over 1 - 3/2} \right\rbrack = {1 \over 4}\left\lbrack - \Psi\left(1 \over 2\right) + 2\Psi\left(1\right) - \Psi\left(3 \over 2\right) \right\rbrack \\[3mm]&= {1 \over 4}\left\lbrace -\Psi\left(1 \over 2\right) + 2\Psi\left(1\right) - \left\lbrack 2 + \Psi\left(1 \over 2\right)\right\rbrack \right\rbrace = {1 \over 2}\left\lbrack -\Psi\left(1 \over 2\right) + \Psi\left(1\right) - 1 \right\rbrack \\[3mm]&= {1 \over 2}\left\lbrace -\left\lbrack-\gamma - 2\ln\left(2\right)\right\rbrack + \left(-\gamma\right) - 1 \right\rbrace = \ln\left(2\right) - {1 \over 2} \\[1cm]& \end{align}

\begin{align} \sum_{n = 0}^{\infty}{1 \over \left(2n + 1\right)\left(2n + 2\right)\left(2n + 3\right)} &= \ln\left(2\right) - {1 \over 2} \\[3mm] \color{#0000FF}{\large\sum_{n = 1}^{\infty}{1 \over \left(2n + 1\right)\left(2n + 2\right)\left(2n + 3\right)}} &= \left\lbrack\ln\left(2\right) - {1 \over 2}\right\rbrack - {1 \over 6} = \color{#0000ff}{\large \ln\left(2\right) - {2 \over 3}} \\[1cm]& \end{align}

$\Psi\left(z\right)$ is the Digamma function and $\gamma = 0.57721566490153286060651209008240243104215933593992\ldots$ is the Euler-Mascheroni constant.