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I need to evaluate the sum given by:

$$\displaystyle \sum_{k=1}^\infty \left( \frac{1}{6k+1}+\frac{1}{6k+3}+\frac{1}{6k+5}-\frac{1}{8k}-\frac{1}{8k+2}-\frac{1}{8k+4}-\frac{1}{8k+6} \right)$$

I know that:

for $k = 1$ I get: $\frac{1}{7}+\frac{1}{9}+\frac{1}{11}-\frac{1}{8}-\frac{1}{10}-\frac{1}{12}\mathbf{-\frac{1}{14}}$

for $k = 2$ I get: $\frac{1}{13}+\frac{1}{15}+\frac{1}{17}-\frac{1}{16}-\frac{1}{18} \mathbf{-\frac{1}{20}-\frac{1}{22}}$

for $k = 3$ I get: $\frac{1}{19}+\frac{1}{21}+\frac{1}{23}-\frac{1}{24} \mathbf{-\frac{1}{26}-\frac{1}{28}-\frac{1}{30}}$

for $k = 4$ I get: $\frac{1}{25}+\frac{1}{27}+\frac{1}{29} \mathbf{ -\frac{1}{32}-\frac{1}{34}-\frac{1}{36}-\frac{1}{38}}$

I can write that sum for $k = 4$ as: $$ \frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}+\frac{1}{13} -\frac{1}{14}+\frac{1}{15}-\frac{1}{16}+\frac{1}{17}-\frac{1}{18}+\frac{1}{19}-\frac{1}{20}+\frac{1}{21}-\frac{1}{22}+\frac{1}{23}-\frac{1}{24}+\frac{1}{25}-\frac{1}{26}+\frac{1}{27}-\frac{1}{28}+\frac{1}{29} -\frac{1}{30} \mathbf{ -\frac{1}{32}-\frac{1}{34}-\frac{1}{36}-\frac{1}{38}}$$

And I can write that sum for $k = 8$ as: $$ \frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}+\frac{1}{13} -\frac{1}{14}+\frac{1}{15}-\frac{1}{16}+\frac{1}{17}-\frac{1}{18}+\frac{1}{19}-\frac{1}{20}+\frac{1}{21}-\frac{1}{22}+\frac{1}{23}-\frac{1}{24}+\frac{1}{25}-\frac{1}{26}+\frac{1}{27}-\frac{1}{28}+\frac{1}{29} -\frac{1}{30} + \frac{1}{31}-\frac{1}{32}+\frac{1}{33}-\frac{1}{43}+\frac{1}{35}-\frac{1}{36}+\frac{1}{37} -\frac{1}{38}+\frac{1}{39}-\frac{1}{40}+\frac{1}{41}-\frac{1}{42}+\frac{1}{43}-\frac{1}{44}+\frac{1}{45}-\frac{1}{46}+\frac{1}{47}-\frac{1}{48}+\frac{1}{49}-\frac{1}{50}+\frac{1}{51}-\frac{1}{52}+\frac{1}{53} -\frac{1}{54} \mathbf{ -\frac{1}{56}-\frac{1}{58} -\frac{1}{60}-\frac{1}{62}-\frac{1}{64}-\frac{1}{66}-\frac{1}{68} -\frac{1}{70}}$$

I see that for every k I get $1$ extra negative element at the end of the sum. I had an idea to rewrite it that way: $$\displaystyle \sum_{k=1}^\infty \left( \frac{1}{6k+1}+\frac{1}{6k+3}+\frac{1}{6k+5}-\frac{1}{8k}-\frac{1}{8k+2}-\frac{1}{8k+4}-\frac{1}{8k+6} \right) = $$ $$\displaystyle \sum_{k=1}^\infty \left( (-1)^{n+1}\frac{1}{n} \right) -1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6} - \displaystyle \sum_{k=1}^\infty X$$

That way I know that the first sum converges (Dirichlet), but still I don't know how to evaluate that expression. I don't know how to evaluate $\displaystyle \sum_{k=1}^\infty ( (-1)^{n+1}\frac{1}{n}$ ) and I don't know how to include those negative elements in my sum (those are marked as X).

Martin R
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mymathc
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    For the first sum, $\sum\limits_{n\ge1}\frac{(-1)^{n+1}}n$: are you aware of $\sum\limits_{n\ge0}x^n=\frac1{1-x}$ for $|x|<1$? If so, consider the integral and what you get for $x=-1$. – user170231 Jan 05 '21 at 18:20
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    See this and this and use the same technique. – V.G Jan 05 '21 at 18:25
  • @user170231, Unfortunately, I can not use integrals yet, thank you for your answer. – mymathc Jan 05 '21 at 18:34
  • @Light Yagami Both those examples have same factor next to variable so I don't find them useful in my case. – mymathc Jan 05 '21 at 18:57
  • @mymathc: Don't forget to put parenthesis around the terms which should be within the scope of the $\Sigma$ symbol. – Markus Scheuer Jan 05 '21 at 19:08
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    @Markus Scheuer Got that fixed. Now it should be clearer. Thank you. – mymathc Jan 05 '21 at 19:14
  • @mymathc: For the alternating harmonic series, read answers here – V.G Jan 05 '21 at 19:26
  • @Light Yagami thank you, I will check that in a moment. However the problem is, now I am not sure if I did it right even to that point. I mean, it may be impossible to evaluate X without that alternating harmonic series. – mymathc Jan 05 '21 at 19:34

2 Answers2

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If you are not allowed to use integrals which the standard method of invoking Abel's Theorem requires, then pretty much all you can do is get creative. Denote:

$$S_n = \sum_{k=0}^n\left(\frac{1}{6k+1} + \frac{1}{6k+3} + \frac{1}{6k+5}\right)$$ and $$T_n = \sum_{k=0}^n\left(\frac{1}{6k+2} + \frac{1}{6k+4} + \frac{1}{6k+6}\right)$$ for starters. Then, you can see that $S_n+T_n = H_{6n+6}$, the usual Harmonic sum. Note that you have to consider these finite truncates, for otherwise each of $S_n$ and $T_n$ diverge.

On the other hand, consider the remaining term in your original sum:

$$R_n = \sum_{k=1}^n\left(\frac{1}{8k} + \frac{1}{8k+2} + \frac{1}{8k+4} + \frac{1}{8k+6}\right)$$

$$= \frac{1}{2}\sum_{k=1}^n\left(\frac{1}{4k}+\frac{1}{4k+1}+\frac{1}{4k+2}+\frac{1}{4k+3}\right) = \dfrac{H_{4n+3} -1-\frac 12-\frac 13 }{2} = \dfrac{H_{4n+3}}{2} - \frac{11}{12}.$$

Now, we just need to "evaluate" $T_n$ nicely:

$$T_n = \frac{1}{2}\sum_{k=0}^n\left(\frac{1}{3k+1} + \frac{1}{3k+2} + \frac{1}{3k+3}\right) = \dfrac{H_{3n+3}}{2}.$$ Therefore, the $n$ -th partial sum of your limit is then:

$$S_n-R_n - \left(1+\frac 13+\frac 15\right) = H_{6n+6} - \dfrac{H_{3n+3}}{2} - \frac{23}{15}-\dfrac{H_{4n+3}}{2}+\frac{11}{12} = $$ $$ \sim \ln(6n+6) - \ln(3n+3)/2-\ln(4n+3)/2-\dfrac{37}{60} = \ln\sqrt{\dfrac{12n+12}{4n+3}}-\dfrac{37}{60}$$ so the answer is: $$\lim\limits_{n\to\infty}\left(\ln\sqrt{\dfrac{12n+12}{4n+3}}-\dfrac{37}{60}\right) = \frac{1}{2}\ln 3 - \frac{37}{60}$$

John Bentin
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dezdichado
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  • Love it that you took into account my constraints. I will read it in a moment and let you know if I see any flaws. Then I will mark my question as answered if everything is ok. Thank you. – mymathc Jan 05 '21 at 19:39
  • @LightYagami it's not - it's continued on the second line. – dezdichado Jan 05 '21 at 19:43
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    Look good to me, thank you once agin, I wouldn't think of it myself. – mymathc Jan 05 '21 at 19:53
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    fewer terms to limit if we take the sums up to $n-1;$ $H_{6n} - \frac{1}{2} H_{3n} - \frac{1}{2} H_{4n-1} - \frac{37}{60} $ For that matter $H_{6n} - \frac{1}{2} H_{3n} - \frac{1}{2} H_{4n} - \frac{37}{60} $ differs jus by shrinking terms and has the same limit – Will Jagy Jan 05 '21 at 23:27
  • Could you just tell me why $H_{6n+6} ~ln(6n+^)$? @dezdichado – mymathc Jan 09 '21 at 00:06
  • @mymathc it is well-known and an easy analysis exercise that: $$\ln n + 1>H_n > \ln(n+1)$$ and so $H_n \sim \ln n$ asymptotically – dezdichado Jan 09 '21 at 01:06
  • Ok, now I get it. I mean the solution is probably right but I ma not sure if I can use it myself. Do you think that there might be a solution that does not require that approximation? I am on a very basic level and because of that thought of using alternating harmonic series, but then it is problematic that there are different factors next to variable. – mymathc Jan 09 '21 at 01:33
  • I think that the whole idea of that excercise might be about rearrangement of elements in alternating harmonic series. That way the value changes as well. – mymathc Jan 09 '21 at 01:45
  • Yeah, now I am pretty sure that rearrangement of elements in alternating harmonic series is what I ma looking for. I will do some research on that! – mymathc Jan 09 '21 at 01:55
  • You can prove the above inequality only using induction and calculus and i am afraid that's as elementary as it gets – dezdichado Jan 09 '21 at 02:09
  • But how do I use induction if I don't know the sum? I could use it if I knew the answer and then needed to prove that. Ahh, you mean only that nequality? I am nearly sure that it can by done by using ideas of rearrangement of elements in alternating harmonic series, but I have no idea how to do that and after whole day I think I will just give up. – mymathc Jan 09 '21 at 03:32
  • I think I got it. In a book on math analysis I found a mathematical formula for that special case of rearrangement of elements in alternating harmonic series where after $n$ positive elements there are $m$ negative elements. I can show you later if you are interested. – mymathc Jan 09 '21 at 03:49
  • It is perfectly fine to write up an answer to your own question should you think it is of value to others. – dezdichado Jan 09 '21 at 04:59
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Hoping that you are aware of the digamma function and harmonic numbers, we can even approximate the partial sums before going to the limit.

Let

$$a_k= \frac{1}{6k+1}+\frac{1}{6k+3}+\frac{1}{6k+5}-\frac{1}{8k}-\frac{1}{8k+2}-\frac{1}{8k+4}-\frac{1}{8k+6}$$ $$S_p=\sum_{k=1}^p a_k=\frac{1}{6} \psi \left(p+\frac{7}{6}\right)+\frac{1}{6} \psi \left(p+\frac{3}{2}\right)+\frac{1}{6} \psi \left(p+\frac{11}{6}\right)-$$ $$\frac{1}{2} \psi (4 p+4)-\frac{\gamma }{2}+\frac{11}{12}-\frac{\psi \left(\frac{11}{6}\right)}{6}-\frac{\psi \left(\frac{3}{2}\right)}{6}-\frac{\psi \left(\frac{7}{6}\right)}{6}$$ Switching to harmonic numbers $$S_p=\frac{1}{2} H_{3 p+\frac{5}{2}}-\frac{1}{2}H_{4 p+3}-\frac{37}{60}+\log (2)$$ Using the asymptotics of harmonic numbers $$S_p=\left(\frac{\log (3)}{2}-\frac{37}{60}\right)+\frac{1}{16 p}-\frac{199}{3456 p^2}+O\left(\frac{1}{p^3}\right)$$

Using this truncated expansion, we would hve $S_{10}=-0.06169$ while the exact value would be $-0.06164$.

Claude Leibovici
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