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How can I find the sum of the following series?

$ S_n = \frac{1}{2 \times 3 \times 4} + \frac{1}{5 \times 6 \times 7} + ...+ \frac{1}{(3n-1) \times 3n \times (3n+1)} $

S.C.B.
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Rajan saha Raju
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    What have you tried? – LHF Mar 13 '20 at 09:25
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    For a general method see https://math.stackexchange.com/q/3538626/631742 – Maximilian Janisch Mar 13 '20 at 09:27
  • Note: $$2S_n=\frac{1}{2}-\frac{2}{3}+\frac{1}{4}+\frac{1}{5} -\frac{2}{6}+\frac{1}{7}\cdots$$ I think you could probably use the series expansion of $\ln x$ to answer your question. Or maybe not. – S.C.B. Mar 13 '20 at 09:30
  • How you found the value of $2 S_n$ – Rajan saha Raju Mar 13 '20 at 09:33
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    $$\frac{2}{n(n+1)(n+2)}=\frac{1}{n}-\frac{2}{n+1}+\frac{1}{n+2}$$ through algebraic manipulation. – S.C.B. Mar 13 '20 at 09:34
  • Yeah! partial fraction. – Rajan saha Raju Mar 13 '20 at 09:37
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    I think that the $n$-th term is $\frac{1}{(3n-1)3n(3n+1)}$, not $\frac{1}{n(n+1){n+2)}$, if you observe the first few terms. – S.C.B. Mar 13 '20 at 09:44
  • @S.C.B., Why did you change the meaning of OP's question? – LHF Mar 13 '20 at 09:45
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    @Atticus The fist few terms were $\frac{1}{2.3.4}$, $\frac{1}{5.6.7}$ and so on, not $\frac{1}{234}$, $\frac{1}{345}$, and so on. I thought that was clear the OP's intention given the first two terms. – S.C.B. Mar 13 '20 at 09:46
  • Yeah. $\frac{1}{(3n-1)(3n-2)(3n-3)}$ is ok – Rajan saha Raju Mar 13 '20 at 09:48
  • @SCB, but how did you know he meant $\frac{1}{2\cdot 3\cdot 4}+\frac{1}{5\cdot 6\cdot 7}+\ldots+\frac{1}{(3n-1)\cdot 3n\cdot (3n+1)}$ and not $\frac{1}{2\cdot 3\cdot 4}+\frac{1}{3\cdot 4\cdot 5}+\ldots+\frac{1}{n\cdot (n+1)\cdot (n+2)}$? – LHF Mar 13 '20 at 09:49
  • @Atticus Looking at the edit history, you can see that the first few terms given was equal to the first series, not the second one. – S.C.B. Mar 13 '20 at 09:50
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    @S.C.B., your edit is probably correct. What I'm saying is OP might have mistaken the second term, not the last. I think the correct action in this case is to ask the OP for clarification, not edit. – LHF Mar 13 '20 at 09:52
  • @Atticus, S.C.B is ok – Rajan saha Raju Mar 13 '20 at 09:52
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    So $\frac{1}{(3n-1)(3n)(3n+1)} = \frac{1}{2(3n-1)} - \frac{1}{3n} + \frac{1}{2(3n+1)}$ and we can convert it into telescoping series . – Aditya Mar 13 '20 at 09:55
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    @Atticus You may be correct. I did leave a comment beforehand-but, I thought the mistake was rather obvious and may not have given the other option proper thought. I confess I haven't used this site for several years, and I don't exactly remember the proper codes of conduct in this site. But I think I do recall that editing rather "obvious looking mistakes" were a bit of a contentious thing on this site, particularly because of ambiguity and whether or not it really is obvious. – S.C.B. Mar 13 '20 at 09:57

4 Answers4

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I'm not sure if there exists an answer for the sum of the first few terms, but for the infinite sum there is an answer. I thought the "series" in your question meant an infinite sum, like here, so sorry if this does not answer your question.

Note $$-\ln(1-x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\cdots$$ Thus $$\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\cdots=-\ln(1- x)-x$$ Putting $x= \exp\left(\frac{2\pi}{3}\right)$, $x= \exp\left(\frac{4\pi}{3}\right)$, and summing the results, we get $$-\frac{1}{2}+\frac{2}{3}-\frac{1}{4}+\cdots=-\ln\left(\left(1-\exp\left(\frac{2\pi}{3}\right) \right)\left(1-\exp\left(\frac{4\pi}{3}\right) \right)\right)+1\tag{*}$$

To calculate $(*)$, note $\left(x-\exp\left(\frac{2\pi}{3}\right) \right)\left(x-\exp\left(\frac{4\pi}{3}\right)\right)=x^2+x+1$, and put $x=1$.

Multiplying by $-1$ and dividing by $2$, we get that the desired sum $S_n$ is equal to $\dfrac{\ln3 -1}{2}$.

S.C.B.
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I used Maple to get $$ \sum _{k=1}^{n}{\frac {1}{ \left( 3\,k-1 \right) (3k) \left( 3\,k+1 \right) }}=\frac16\,\psi \left( n+\frac23 \right) -\frac13\,\psi \left( n+1 \right) +\frac16\,\psi \left( n+\frac43 \right) +\frac12\,\ln \left( 3 \right) - \frac12 $$ Here $\psi$ is the digamma function $\psi(x) = \frac{\Gamma'(x)}{\Gamma(x)}$

GEdgar
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Use partial fractions. This decomposes into a collection of harmonic sums, which you can combine/simplify.

vonbrand
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As said in comments and answers, using partial fraction decomposition and harmonic numbers $$S_n=\frac{1}{6}H_{n-\frac{1}{3}}-\frac{1}{3}H_n+\frac{1}{6}H_{n+\frac{1}{3}}+ \frac{\log (3)-1}{2}$$ Uing asymptotics $$S_n=\frac{\log (3)-1}{2}-\frac{1}{54 n^2}+\frac{1}{54 n^3}+O\left(\frac{1}{n^4}\right)$$ Use your pocket calculator for $n=5$; the exact result is $$S_5=\frac{14039}{288288}\approx 0.048698$$ while the above truncated expansion gives $$\frac{\log (3)}{2}-\frac{3379}{6750}\approx 0.048713$$

Claude Leibovici
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