4
\documentclass{article}

\usepackage{amsmath}

\begin{document}

\begin{align*}
\sum_{k=1}^\infty \frac{1}{k} &\rightarrow \infty
&\sum_{k=1}^\infty \frac{1}{k^2} &= \frac{\pi^2}{6}
&&\sum_{k=1}^\infty \frac{1}{k^3} = \zeta(3)
\\
\\
\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} &= \ln(2)
&\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^2} &= \frac{\pi^2}{12}
&&\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^3} = \frac{3}{4}\zeta(3)
\\
\\
\sum_{k=1}^\infty \frac{1}{2k} &\rightarrow \infty
&\sum_{k=1}^\infty \frac{1}{{(2k)}^2} &= \frac{\pi^2}{24}
&&\sum_{k=1}^\infty \frac{1}{{(2k)}^3} = \frac{1}{8}\zeta(3)
\\
\\
\sum_{k=1}^\infty \frac{(-1)^{k+1}}{2k} &= \frac{\ln(2)}{2}
&\sum_{k=1}^\infty \frac{(-1)^{k+1}}{{(2k)}^2} &= \frac{\pi^2}{48}
&&\sum_{k=1}^\infty \frac{(-1)^{k+1}}{{(2k)}^3} = \frac{3}{32}\zeta(3)
\\
\\
\sum_{k=0}^\infty \frac{1}{2k+1} &\rightarrow \infty
&\sum_{k=0}^\infty \frac{1}{{(2k+1)}^2} &= \frac{\pi^2}{8}
&&\sum_{k=0}^\infty \frac{1}{{(2k+1)}^3} = \frac{7}{8}\zeta(3)
\\
\\
\sum_{k=0}^\infty \frac{(-1)^{k}}{2k+1} &= \frac{\pi}{4}
&\sum_{k=0}^\infty \frac{(-1)^{k}}{{(2k+1)}^2} &\neq \frac{\pi^2}{16}
&&\sum_{k=0}^\infty \frac{(-1)^{k}}{{(2k+1)}^3} = \frac{\pi^3}{32} \neq \frac{21}{32}\zeta(3)
\end{align*}

\end{document}

Sorry, I don't know how to display the output of this code, so I uploaded the pdf image. I am trying to get the sums to align like in the last column. I think it would look better if both the sums and the equals signs are aligned though. sums

Werner
  • 603,163
zerosofthezeta
  • 249

4 Answers4

3

alignat is best suited for this case. Remember that alignat alternates alignments left then right and so on, so you should add double && to get left alignment.

\documentclass{article}
\usepackage{amsmath}
\begin{document}

\begin{alignat*}{3}
&\sum_{k=1}^\infty \frac{1}{k} \rightarrow \infty
&&\sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6}
&&\sum_{k=1}^\infty \frac{1}{k^3} = \zeta(3)
\\
\\
&\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} = \ln(2)
&&\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^2} = \frac{\pi^2}{12}
&&\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^3} = \frac{3}{4}\zeta(3)
\\
\\
&\sum_{k=1}^\infty \frac{1}{2k} \rightarrow \infty
&&\sum_{k=1}^\infty \frac{1}{{(2k)}^2} = \frac{\pi^2}{24}
&&\sum_{k=1}^\infty \frac{1}{{(2k)}^3} = \frac{1}{8}\zeta(3)
\\
\\
&\sum_{k=1}^\infty \frac{(-1)^{k+1}}{2k} = \frac{\ln(2)}{2}
&&\sum_{k=1}^\infty \frac{(-1)^{k+1}}{{(2k)}^2} = \frac{\pi^2}{48}
&&\sum_{k=1}^\infty \frac{(-1)^{k+1}}{{(2k)}^3} = \frac{3}{32}\zeta(3)
\\
\\
&\sum_{k=0}^\infty \frac{1}{2k+1} \rightarrow \infty
&&\sum_{k=0}^\infty \frac{1}{{(2k+1)}^2} = \frac{\pi^2}{8}
&&\sum_{k=0}^\infty \frac{1}{{(2k+1)}^3} = \frac{7}{8}\zeta(3)
\\
\\
&\sum_{k=0}^\infty \frac{(-1)^{k}}{2k+1} = \frac{\pi}{4}
&&\sum_{k=0}^\infty \frac{(-1)^{k}}{{(2k+1)}^2} \neq \frac{\pi^2}{16}
&&\sum_{k=0}^\infty \frac{(-1)^{k}}{{(2k+1)}^3} = \frac{\pi^3}{32} \neq \frac{21}{32}\zeta(3)
\end{alignat*}

\end{document}

enter image description here

AboAmmar
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3

One option is given by:

\documentclass[12pt]{article}

\usepackage{amsmath}

\begin{document}

\begin{align*}
&\sum_{k=1}^\infty \frac{1}{k} \rightarrow \infty 
&& \sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6} 
&& \sum_{k=1}^\infty \frac{1}{k^3} = \zeta(3)
\\
&\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} = \ln(2) 
&& \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^2} = \frac{\pi^2}{12}  
&& \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^3} = \frac{3}{4}\zeta(3)
\\
&\sum_{k=1}^\infty \frac{1}{2k} \rightarrow \infty 
&& \sum_{k=1}^\infty \frac{1}{{(2k)}^2} = \frac{\pi^2}{24} 
&& \sum_{k=1}^\infty \frac{1}{{(2k)}^3} = \frac{1}{8}\zeta(3)
\\
& \sum_{k=1}^\infty \frac{(-1)^{k+1}}{2k} = \frac{\ln(2)}{2}
&& \sum_{k=1}^\infty \frac{(-1)^{k+1}}{{(2k)}^2} = \frac{\pi^2}{48}
&& \sum_{k=1}^\infty \frac{(-1)^{k+1}}{{(2k)}^3} = \frac{3}{32}\zeta(3)
\\
& \sum_{k=0}^\infty \frac{1}{2k+1} \rightarrow \infty
&& \sum_{k=0}^\infty \frac{1}{{(2k+1)}^2} = \frac{\pi^2}{8}
&& \sum_{k=0}^\infty \frac{1}{{(2k+1)}^3} = \frac{7}{8}\zeta(3)
\\
& \sum_{k=0}^\infty \frac{(-1)^{k}}{2k+1} = \frac{\pi}{4}
&& \sum_{k=0}^\infty \frac{(-1)^{k}}{{(2k+1)}^2} \neq \frac{\pi^2}{16}
&& \sum_{k=0}^\infty \frac{(-1)^{k}}{{(2k+1)}^3} = \frac{\pi^3}{32} \neq \frac{21}{32}\zeta(3)
\end{align*}

\end{document}

which produces:

enter image description here

Notes: more space can be added between columns by using \quad after the first and second equations of each row. Space can be added between rows by using \\ on two lines as presented by the original code of the proposer.

Leucippus
  • 1,636
2

I propose a classic alignment (with align*), another (with alignat*), perhaps more pleasant to read, which consists in not repeating the sum operator on each line, and a last with a single big sum operator for the whole alignment:

\documentclass{article}

\usepackage{amsmath}
\usepackage{graphics} 
\usepackage{relsize}
\newcommand\Sum{\mathlarger{\sum}}
\usepackage{lettrine} 
\usepackage{showframe} 
\begin{document}

\begin{align*}
    & \sum_{k=1}^\infty \frac{1}{k}\rightarrow \infty &
  &\sum_{k=1}^\infty \frac{1}{k^2}= \frac{\pi^2}{6}
    & & \sum_{k=1}^\infty \frac{1}{k^3} = \zeta(3)
  \\[1ex]
  & \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}= \ln(2)
    & & \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^2}= \frac{\pi^2}{12}
    & & \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^3} = \frac{3}{4}\zeta(3)
  \\[1ex]
  & \sum_{k=1}^\infty \frac{1}{2k} \rightarrow \infty
    & & \sum_{k=1}^\infty \frac{1}{{(2k)}^2} = \frac{\pi^2}{24}
    & & \sum_{k=1}^\infty \frac{1}{{(2k)}^3} = \frac{1}{8}\zeta(3)
  \\[1ex]
  & \sum_{k=1}^\infty \frac{(-1)^{k+1}}{2k}= \frac{\ln(2)}{2}
    & & \sum_{k=1}^\infty \frac{(-1)^{k+1}}{{(2k)}^2} = \frac{\pi^2}{48}
    & & \sum_{k=1}^\infty \frac{(-1)^{k+1}}{{(2k)}^3} = \frac{3}{32}\zeta(3)
  \\[1ex]
  & \sum_{k=0}^\infty \frac{1}{2k+1}\rightarrow \infty
    & & \sum_{k=0}^\infty \frac{1}{{(2k+1)}^2}= \frac{\pi^2}{8}
    & & \sum_{k=0}^\infty \frac{1}{{(2k+1)}^3} = \frac{7}{8}\zeta(3)
  \\[1ex]
  &\sum_{k=0}^\infty \frac{(-1)^{k}}{2k+1}= \frac{\pi}{4}
    & & \sum_{k=0}^\infty \frac{(-1)^{k}}{{(2k+1)}^2}\neq \frac{\pi^2}{16}
    & & \sum_{k=0}^\infty \frac{(-1)^{k}}{{(2k+1)}^3} = \frac{\pi^3}{32} \neq \frac{21}{32}\zeta(3)
\end{align*}

\begin{alignat*}{3}
  \mathlarger{\sum}_{k=1}^\infty \enspace & \frac{1}{k}\rightarrow \infty
    & \quad \Sum_{k=1}^\infty\enspace &
  \frac{1}{k^2}= \frac{\pi^2}{6}
    & \quad \Sum_{k=1}^\infty\enspace & \frac{1}{k^3} = \zeta(3)
  \\[1ex]
  &\frac{(-1)^{k+1}}{k}= \ln(2)
    & & \frac{(-1)^{k+1}}{k^2}= \frac{\pi^2}{12}
    & & \frac{(-1)^{k+1}}{k^3} = \frac{3}{4}\zeta(3)
  \\[1ex]
  & \frac{1}{2k} \rightarrow \infty
    & & \frac{1}{{(2k)}^2} = \frac{\pi^2}{24}
    & & \frac{1}{{(2k)}^3} = \frac{1}{8}\zeta(3)
  \\[1ex]
  & \frac{(-1)^{k+1}}{2k}= \frac{\ln(2)}{2}
    & & \frac{(-1)^{k+1}}{{(2k)}^2} = \frac{\pi^2}{48}
    & & \frac{(-1)^{k+1}}{{(2k)}^3} = \frac{3}{32}\zeta(3)
  \\[1ex]
  & \frac{1}{2k+1}\rightarrow \infty
    & & \frac{1}{{(2k+1)}^2}= \frac{\pi^2}{8}
    & & \frac{1}{{(2k+1)}^3} = \frac{7}{8}\zeta(3)
  \\[1ex]
  & \frac{(-1)^{k}}{2k+1}= \frac{\pi}{4}
    & & \frac{(-1)^{k}}{{(2k+1)}^2}\neq \frac{\pi^2}{16}
    & & \frac{(-1)^{k}}{{(2k+1)}^3} = \frac{\pi^3}{32} \neq \frac{21}{32}\zeta(3)
\end{alignat*}


\begin{alignat*}{4}
\smash{\mathop{\raisebox{-0.7\height}{\scalebox{2.8}[4]{$ \sum $}}}_{\boldsymbol{k=}1}^{\boldsymbol\infty}}\! \enspace
 & \frac{1}{k}\rightarrow \infty
    & \hspace{2.5em} &
  \frac{1}{k^2}= \frac{\pi^2}{6}
    & \hspace{2.5em} & \frac{1}{k^3} = \zeta(3)
  \\
  &\frac{(-1)^{k+1}}{k}= \ln(2)
    & & \frac{(-1)^{k+1}}{k^2}= \frac{\pi^2}{12}
    & & \frac{(-1)^{k+1}}{k^3} = \frac{3}{4}\zeta(3)
  \\[1ex]
  & \frac{1}{2k} \rightarrow \infty
    & & \frac{1}{{(2k)}^2} = \frac{\pi^2}{24}
    & & \frac{1}{{(2k)}^3} = \frac{1}{8}\zeta(3)
  \\[1ex]
  & \frac{(-1)^{k+1}}{2k}= \frac{\ln(2)}{2}
    & & \frac{(-1)^{k+1}}{{(2k)}^2} = \frac{\pi^2}{48}
    & & \frac{(-1)^{k+1}}{{(2k)}^3} = \frac{3}{32}\zeta(3)
  \\[1ex]
  & \frac{1}{2k+1}\rightarrow \infty
    & & \frac{1}{{(2k+1)}^2}= \frac{\pi^2}{8}
    & & \frac{1}{{(2k+1)}^3} = \frac{7}{8}\zeta(3)
  \\[1ex]
  & \frac{(-1)^{k}}{2k+1}= \frac{\pi}{4}
    & & \frac{(-1)^{k}}{{(2k+1)}^2}\neq \frac{\pi^2}{16}
    & & \frac{(-1)^{k}}{{(2k+1)}^3} = \frac{\pi^3}{32} \neq \frac{21}{32}\zeta(3)
\end{alignat*}

\end{document}

enter image description here

enter image description here

AboAmmar
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Bernard
  • 271,350
  • Personally, I like it with the sums included. But this got me thinking... Is there a way to enlarge the sum operator to where it is the size of six rows? Instead of having one regular sized sum on top, is there a way to have one large sum to the left of the rows? (Please tell me you know how to create this!) – zerosofthezeta Sep 04 '15 at 00:01
  • Usually I include the sums, but here they are quite a lot, I feel it tiresome to the eye. I'm not sure I understand very well your request. The second solution makes the sum signs column heads, so to say. Do you mean only one quite big sign instead of three moderately larger? – Bernard Sep 04 '15 at 00:13
  • Imagine the first column alone without the sums. Is there way to enlarge the sum symbol so it is the size of an entire page and place it to the left of the column? So there is only one sum symbol for the 6 equations. The closest analogy I can think of is an enormous drop capital at the beginning of a chapter in a book. – zerosofthezeta Sep 04 '15 at 00:21
  • @zerosofthezeta: I've just added a third possibility. Is it close to what you imagined? – Bernard Sep 04 '15 at 00:52
  • Yes, its close...I think what I am imagining is more of an artistic thing, perhaps not suitable for LaTex...Thanks for your help though! I really appreciate all the options you provided! – zerosofthezeta Sep 04 '15 at 01:56
  • If you can post a scan of a hand-made sketch of what you precisely imagine, I might think about it. I like challenges! – Bernard Sep 04 '15 at 08:53
  • Ok, here you go: http://i.imgur.com/FQQanZZ.png . You love challenges, huh? Me too! Writing $\sum_{k=1}^\infty \frac{1}{k^3}$ in terms of other constants (just like the first sum in the second column) is a 300+ year old math problem that I hope to solve. – zerosofthezeta Sep 04 '15 at 21:58
  • All is know is that it is irational, I believe? I'll try to do something tomorrow, but it is not sensible to have the sum symbol the height of the whole alignment. – Bernard Sep 04 '15 at 23:08
1

You may just as well set the entire construction inside an array:

enter image description here

\documentclass{article}
\usepackage{amsmath,array}

\begin{document}

\[
  \renewcommand{\arraystretch}{3}% http://tex.stackexchange.com/a/31704/5764
  \begin{array}{ >{\displaystyle}l @{\quad} >{\displaystyle}l @{\quad} >{\displaystyle}l }
    \sum_{k=1}^\infty \frac{1}{k} \rightarrow \infty
    & \sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6}
    & \sum_{k=1}^\infty \frac{1}{k^3} = \zeta(3)
    \\
    \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} = \ln(2)
    & \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^2} = \frac{\pi^2}{12}
    & \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^3} = \frac{3}{4}\zeta(3)
    \\
    \sum_{k=1}^\infty \frac{1}{2k} \rightarrow \infty
    & \sum_{k=1}^\infty \frac{1}{{(2k)}^2} = \frac{\pi^2}{24}
    & \sum_{k=1}^\infty \frac{1}{{(2k)}^3} = \frac{1}{8}\zeta(3)
    \\
    \sum_{k=1}^\infty \frac{(-1)^{k+1}}{2k} = \frac{\ln(2)}{2}
    & \sum_{k=1}^\infty \frac{(-1)^{k+1}}{{(2k)}^2} = \frac{\pi^2}{48}
    & \sum_{k=1}^\infty \frac{(-1)^{k+1}}{{(2k)}^3} = \frac{3}{32}\zeta(3)
    \\
    \sum_{k=0}^\infty \frac{1}{2k+1} \rightarrow \infty
    & \sum_{k=0}^\infty \frac{1}{{(2k+1)}^2} = \frac{\pi^2}{8}
    & \sum_{k=0}^\infty \frac{1}{{(2k+1)}^3} = \frac{7}{8}\zeta(3)
    \\
    \sum_{k=0}^\infty \frac{(-1)^{k}}{2k+1} = \frac{\pi}{4}
    & \sum_{k=0}^\infty \frac{(-1)^{k}}{{(2k+1)}^2} \neq \frac{\pi^2}{16}
    & \sum_{k=0}^\infty \frac{(-1)^{k}}{{(2k+1)}^3} = \frac{\pi^3}{32} \neq \frac{21}{32}\zeta(3)
  \end{array}
\]

\end{document}

Padding between columns are set to \quad.

Werner
  • 603,163
  • Interesting. What are the benefits of using an array over align? – zerosofthezeta Sep 03 '15 at 23:55
  • Not really. Your application just didn't call for a align-specific requirement. The entire array block will be unbreakable across the page boundary, while align-like environment allow for breaking in certain situations. – Werner Sep 04 '15 at 00:04