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$$\sum\limits_{n=1}^{\infty} \left(\frac{1}{4 n-3}+\frac{1}{4 n-1}-\frac{1}{2 n}\right) = \;?$$

I have been trying to see if it can be written as sum of two telescope terms but it looks tricky. Any help ?

AlexR
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lilly
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3 Answers3

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Hint:

You might compare the partial sum for $n=1$ to $k$ with $\frac11 -\frac12+\frac13 -\frac14+\cdots -\frac1{4k}$, which has a well known limit.

Added:

If you do, you will find the difference is $\left(\frac12-\frac14\right)+\left(\frac16-\frac18\right)+\cdots+\left(\frac1{4‌​k+2}-\frac1{4k}\right)$ which is half $\frac11 -\frac12+\frac13 -\frac14+\cdots -\frac1{2k}$.

So the limit of the original sum is $$\log_e 2 + \frac12 \log_e 2 \\=\frac32 \,\log_e 2.$$

Henry
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    You can't rearrange terms like this if the series is not absolutely convergent. – TonyK Mar 17 '15 at 09:56
  • @TonyK compare with. Comparison will give a result. – AlexR Mar 17 '15 at 09:56
  • @TonyK: You can rearrange a finite number of terms in a partial sum. – Henry Mar 17 '15 at 09:57
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    @Henry I'm not convinced the partial sum goes out the way you said. For example, exanding out to $n = 2$ gives $(1 + \frac{1}{3} - \frac{1}{2}) + (\frac{1}{5} + \frac{1}{7} - \frac{1}{4})$ – hunter Mar 17 '15 at 09:58
  • @hunter hence compare with. Note that $- \frac{1}{2} = -\frac{1}{4} - \frac{1}{4}$. – MT_ Mar 17 '15 at 10:00
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    how do you prove a series is divergent by comparing to a convergent series ? – hunter Mar 17 '15 at 10:01
  • @hunter The series is convergent. – MT_ Mar 17 '15 at 10:02
  • @hunter: the difference between the partial sums is $\left(\frac12-\frac14\right)+\left(\frac16-\frac18\right)+\cdots+\left(\frac1{4k+2}-\frac1{4k}\right)$, which is half of something familiar – Henry Mar 17 '15 at 10:06
  • @Henry got it, thanks ! – hunter Mar 17 '15 at 10:22
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Hint: The sum is a rearrangement of the terms of the alternating harmonic series.

Can you find a series to compare term-wise to? Note that $-\frac{1}{2n} = - \frac{1}{4n} - \frac{1}{4n}$.

Since everyone is very hectic in the comments, let me be clear here:

The sum converges. The rearrangement itself is not enough to prove this, of course, since the harmonic series converges conditionally. However, my hint that it is a rearrangement was to hint at the possibility that we can use a comparison to prove it.

Full solution

$ \sum \left( \frac{1}{4n-3} + \frac{1}{4n-1} - \frac{1}{4n} - \frac{1}{4n} \right) = \sum \left(\frac{1}{4n-3} - \frac{1}{4n-2} + \frac{1}{4n-1} - \frac{1}{4n}\right) + \sum \left( \frac{1}{4n-2} - \frac{1}{4n}\right).$

$= \ln 2 + \frac{1}{2} \sum \left( \frac{1}{2n-1} - \frac{1}{2n} \right) = \frac{3}{2} \ln 2.$

MT_
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  • but it's somewhat of an infinite/nasty rearrangement, since the odd-denominator terms are much higher than the even-denominator terms in any given partial sum. – hunter Mar 17 '15 at 09:53
  • Yep - this was a very "just for starters" kind of hint. – MT_ Mar 17 '15 at 09:54
  • ah got it. I guess our comments give a second hint then. – hunter Mar 17 '15 at 09:54
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    Yes I have noticed that and since it is n rearrangement of a conditionally convergent series, it need not converge to $\ln 2$ right – lilly Mar 17 '15 at 09:54
  • note $-\frac1{2n} = -\frac1{4n}-\frac1{4n}$ – Henry Mar 17 '15 at 09:54
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    @lilly In fact using the hint you can write out the difference between $$\sum_{n=1}^k \frac1{4n-3} + \frac1{4n-1} - \frac1{2n} - \sum_{n=1}^l \frac{(-1)^k}k$$ An see that it grows like the harmonic series. – AlexR Mar 17 '15 at 09:56
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    @lilly: That is absolutely right. The rearrangement is not allowed in this case. – TonyK Mar 17 '15 at 09:56
  • @TonyK My mentioning of the rearrangement is that it makes a next possible step of comparison perhaps more natural. – MT_ Mar 17 '15 at 09:59
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    Let's be clear here -- I think @Soke is on team "it diverges" and views this hint as a first step to prove that. But the hint could be misread as "it converges because it can be rearranged." (I am on team "it diverges.") – hunter Mar 17 '15 at 10:00
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    @hunter No, I am on team "it converges," and I am mentioning the rearrangement to give OP some insight on a possible comparison to prove this. – MT_ Mar 17 '15 at 10:01
  • @Soke got it, finally! – hunter Mar 17 '15 at 10:11
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$$\sum\limits_{n=1}^k \left(\frac{1}{4 n-3}+\frac{1}{4 n-1}-\frac{1}{2 n}\right) =1-\frac12+\frac13-\frac14+\cdots-\frac{1}{2k} + \sum\limits_{n=k}^{2k-1} \frac{1}{2n+1}$$

Can you evaluate the limits of the two series on the right-hand side?

TonyK
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