14
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ Welcome to MSE! Have you tried anything and can share your thoughts and attempts? Regards $\endgroup$ – Amzoti Apr 20 '13 at 23:52
  • $\begingroup$ 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. $\endgroup$ – Felix Marin Aug 26 '13 at 9:46
17
$\begingroup$

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 $f(1)$.

$\endgroup$
  • 1
    $\begingroup$ +1. This is a nice and probably an easier approach than mine, though the reason why both work are similar if not the same. $\endgroup$ – user17762 Apr 21 '13 at 0:15
  • $\begingroup$ 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... $\endgroup$ – Sharkos Apr 21 '13 at 0:17
13
$\begingroup$

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.

$\endgroup$
12
$\begingroup$

This is similar, but without partial fractions!

$$(1)$$

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

$$(2)$$

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

$\endgroup$
  • $\begingroup$ +1. This is neat. Slightly circumvents the possible mistake of rearranging sums, that could happen in my method. $\endgroup$ – user17762 Apr 21 '13 at 1:34
  • $\begingroup$ @user17762 Thank you! $\endgroup$ – L. F. Apr 21 '13 at 1:37
6
$\begingroup$

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


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

$\endgroup$
  • $\begingroup$ I am not completely sure but looks like you are rearranging terms for the second summation (?) $\endgroup$ – user17762 Apr 21 '13 at 1:47
  • $\begingroup$ @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. $\endgroup$ – robjohn Apr 21 '13 at 3:34
  • $\begingroup$ 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 $\infty$ (?). $\endgroup$ – user17762 Apr 21 '13 at 3:37
  • 1
    $\begingroup$ @user17762: I was too exuberant with my sigmas. It should have been one sum broken over two lines. $\endgroup$ – robjohn Apr 21 '13 at 9:39
-1
$\begingroup$

\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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.