I need to calculate the sum of the series:$T_{3n}=$ $1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}++-...$ I know that $T_{3n}=\sum_{k=1}^{n}\frac{1}{4k-3}+\frac{1}{4k-1}-\frac{1}{2k}$. And they gave a hint that $u_n=S_n-I_n$ , where $S_n=\sum_{k=1}^n \frac{1}{k}$ and $I_n=\log(n)$, converges. Can anyone give me a direction? I've tried to write $T_{3n}$ in terms of $S_n$ but without success.

  • $\begingroup$ Ah, so the general pattern is odd, odd, even (subtracting)? $\endgroup$ – Simply Beautiful Art Dec 24 '16 at 16:05
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    $\begingroup$ Look at $S_{4n} - T_{3n}$. $\endgroup$ – Daniel Fischer Dec 24 '16 at 16:07
  • $\begingroup$ What is the series? Is $T_{3n} = \sum $ a definition or something deduced? If $T_{3n} = \sum$ what the heck are $T_{3n-2}$ and $T_{3n -1}$? What is ++- supposed to be. $\endgroup$ – fleablood Dec 24 '16 at 16:15
  • $\begingroup$ Okay so if $T_n = \sum_{i=1}^n a_i$ then $a_{3n} = -\frac 1{2n}; a_{3n+1} \frac 1{4n+1}; a_{3n+1} = \frac 1{4n+3}$? $\endgroup$ – fleablood Dec 24 '16 at 16:28

Hint: Add two hopefully known converging series term-wise:

$$ \begin{array}{cccccccccccc} &1&-&\dfrac12&+&\dfrac13&-&\dfrac14&+&\dfrac15&-&\dfrac16&+&\dfrac17&-&\dfrac18&+&\dfrac19&-&\dfrac1{10}&+&\cdots\\ \\ &0&+&\dfrac12&+&0&-&\dfrac14&+&0&+&\dfrac16&+&0&-&\dfrac18&+&0&+&\dfrac1{10}&+&\cdots\\ \\ \hline \\ =&1&+&0&+&\dfrac13&-&\dfrac12&+&\dfrac15&+&0&+&\dfrac17&-&\dfrac14&+&\dfrac19&+&0&+&\cdots \end{array} $$

  • $\begingroup$ Hi, did you mean something like that?: $\sum_{n=0}^\infty \frac{(-1)^nx^n}{n}+\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{2n}$=$ln(1+x)+\frac12ln(1+x)=_{(x=1)}\frac32ln(2)$ $\endgroup$ – CodeHoarder Dec 24 '16 at 21:08
  • $\begingroup$ Yup, @CodeHoarder. That was the idea. Provided that you already know that your series sums to $\ln2$, when $x=1$. That fact is often covered, but it isn't entirely trivial. $\endgroup$ – Jyrki Lahtonen Dec 24 '16 at 23:32
  • $\begingroup$ @CodeHoarder: Also take a look at this. $\endgroup$ – Jyrki Lahtonen Dec 25 '16 at 16:10
  • $\begingroup$ There's a little problem. There's a theorem that basically says that you can't change the order of an infinite sum unless it absolutely converges, so i need to show that the power series i've shown here converge absolutely. $\endgroup$ – CodeHoarder Dec 27 '16 at 21:49
  • $\begingroup$ @CodeHoarder That's exactly the point here (the end result has the terms of the first series but in different order). And if you look carefully you see that I am not doing anything like changing the order of the summation. I am adding two convergent series term-by-term. And that is something we are allowed to do even to conditionally convergent series. $\endgroup$ – Jyrki Lahtonen Dec 27 '16 at 22:43

Let $H_n=\sum\limits_{k=1}^n\frac{1}{k}$ Then as is well known $H_n-\ln n$ converges to $\gamma$

Now $$T_n=H_{4n-1}-\frac{1}{2}H_{2n-1}-\frac{1}{2}H_n$$ thus

$$T_n-\ln (4n-1)+\frac{1}{2}\ln (2n-1)+\frac{1}{2}\ln n \to 0$$ and

$$T_n\to \frac{3}{2}\ln 2$$

  • $\begingroup$ This answer seems to be at variance with the accepted answer. But I don't see an error in my calculations. $\endgroup$ – Rene Schipperus Dec 24 '16 at 16:35
  • $\begingroup$ I just saw that answer first. And i don't quite understand this one. $\endgroup$ – CodeHoarder Dec 24 '16 at 16:44
  • $\begingroup$ Yeah I made some error also, hope its correct now, gotta have dinner now, merry Christmas. $\endgroup$ – Rene Schipperus Dec 24 '16 at 17:11
  • $\begingroup$ Thanks, it's all clear now, merry Christmas $\endgroup$ – CodeHoarder Dec 24 '16 at 17:12

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