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Determine whether the difference of the following two series is convergent or not and Prove your answer$$ \sum_{n=1}^\infty \frac{1}{n} $$ and $$\sum_{n=1}^\infty \frac{1}{2n-1} $$

What i tried. I said that the difference of the two series is divergent. My proof is as follows. Find the difference of the two series to get $$\sum_{n=1}^\infty \frac{1}{n} -\sum_{n=1}^\infty \frac{1}{2n-1} = \sum_{n=1}^\infty \frac{n-1}{n(2n-1)}$$ But this diffcult to prove directly that $\sum_{n=1}^\infty \frac{n-1}{n(2n-1)}$ is divergent. So i tired proving by contradiction by assuming that it is convergent and by rearraging the above equation we have,$\sum_{n=1}^\infty \frac{1}{n} =\sum_{n=1}^\infty \frac{1}{2n-1} + \sum_{n=1}^\infty \frac{n-1}{n(2n-1)}$ and since $\sum_{n-1}^\infty \frac{n-1}{n(2n-1)}$ is convergent by our assumption and $\sum_{n=1}^\infty \frac{1}{2n-1}$ is also convergent (need to be proven) then the sum of both series also have to be convergent and thus contradicting the fact that $\sum_{n=1}^\infty \frac{1}{n}$ is divergent and thus proving the statement. Is my proof correct and is there a better prove. Could anyone explain the Prove to me. Thanks

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  • $\begingroup$ Sum $1/(2n+1)$ is certainly divergent so it isn't correct, though why not do a comparison test on the serie you want? $\endgroup$
    – kingW3
    Apr 7, 2017 at 11:15
  • $\begingroup$ Well i tried all the different tests and apparently it dosent work for this question. $\endgroup$
    – ys wong
    Apr 7, 2017 at 11:18
  • $\begingroup$ Heuristically, for large $n$, $\frac{n-1}{n(2n-1)}$ is the same as $\frac{n}{2n^2}=\frac{1}{2n}$, which is divergent. Apply the comparison test to acutally prove it. $\endgroup$ Apr 7, 2017 at 11:19
  • $\begingroup$ @NickyHekster ratio test doesn't work $\endgroup$ Apr 7, 2017 at 11:20
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    $\begingroup$ Just a sidenote. You are not allowed rearrange divergent sums. This can go terribly wrong. You need to work with the partial sums and then use the partial sums to argue if the difference does converge or not.You should carefully ready your exercise. $\endgroup$
    – MrYouMath
    Apr 7, 2017 at 12:00

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A few things need correction in your proof. First, let's compliment you for the idea of trying the contrapositive. Unfortunately, the idea falls flat due to a mistake:

$\sum_{n=1}^\infty \frac{1}{2n-1}$ is not convergent!

To do the question, we need an observation about the difference. First, write down the first few terms of the sum: $$ \sum_{n=1}^\infty \left(\frac 1{2n-1}\right) = 1 + \frac 13 + \frac 15 + \frac 17 + ... $$

you can see that we are precisely summing over the reciprocals of odd numbers here.

In the series $\sum \frac 1n$, we are summing over reciprocals of all numbers. Hence, the difference of these two series, is the series whose terms are the reciprocals of all even numbers. This should be clear.

Once this is clear, we see that every even number is of the form $2k$, where $k$ is a natural number, and vice versa, so that: $$ \sum \frac 1n - \sum \frac 1{2n- 1} = \sum \frac{1}{2n} $$

(Note that the above has nothing to do with convergence of any series, since we are adding series it is mathematically sound).

Now, note that $\sum \frac{1}{2n}$ cannot converge, for if it did, say the sum is $M$, then $\sum \frac 1n = 2M$, which is a contradiction as this series doesn't converge.

Hence, the difference does not converge.

To see that the reciprocals of odd numbers does not converge, just do: $$ 1 + \frac 13 + \frac 15 + \frac 17... > \frac 12 + \frac 14 + \frac 16 + \frac 18... $$

(Compare term by term)

Hence, we have that by the comparison test this is also not convergent.


EDIT : As the below comments point out, a rearrangement of the series $\sum \frac 1n - \sum \frac1{2n-1}$ may converge. However, the original problem also does not point this out explicitly, and in that case it is with me to decide which rearrangement is being referred to : and from the computations made above (where the difference was taken explicitly), I concluded that there was no rearrangement of the terms done while taking the difference. Nevertheless, this is a non-trivial thing that was not pointed out in the question.

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    $\begingroup$ You have to be a little more careful when subtracting series. $\sum\frac1n-\sum\frac1{2n-1}=\sum\frac1{2n}$ is only true for a certain way of looking at the partial sums. But it's also possible to arrange the partial sums to get a convergent sequence. The original problem also has this flaw, but things are ok as long as you mean $\sum_{n=1}^N\frac1n-\frac1{2n-1}$. But as soon as you allow the partial sums to have different upper limits, all bets are off. $\endgroup$
    – Teepeemm
    Apr 7, 2017 at 14:43
  • $\begingroup$ @PaulSinclair: If we're talking about formal series, then it's not clear that the conclusion has anything to do with the question. It's especially muddied because it's not even taking the termwise difference. AFAIK it's actually a nontrivial problem to come up with a self-consistent way of formally manipulating series that allow manipulations such as the one used. $\endgroup$
    – user14972
    Apr 7, 2017 at 15:50
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    $\begingroup$ @PaulSinclair: For example, using the same rule "take two from the first and one from the second" for subtracting the series gives both $$\sum \frac{1}{n} - \sum \frac{1}{2n-1} = \sum \frac{1}{2n} $$ $$\sum \frac{1}{2n-1} - \sum \frac{1}{n} = \sum \frac{1}{4n^3 - n}$$ The latter sum, incidentally, being a convergent one. $\endgroup$
    – user14972
    Apr 7, 2017 at 16:11
  • $\begingroup$ @Hurkyl - mea culpa - when I read the answer, I knew what he meant and saw that it was acceptable for the heuristic understanding he wanted. But by the time I saw the comment, I seem to have mixed up why it was acceptable. My apologies to Teepeemm. I've deleted the offending comment. $\endgroup$ Apr 7, 2017 at 23:23
  • $\begingroup$ Ok, I see what is going on in here. All right, I shall make the appropriate edit. Thank you, @Teepeemm for the clarification. $\endgroup$ Apr 8, 2017 at 3:11
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For a direct comparison test, note that $$ \frac{(n-1)}{n(2n - 1)} \geq \frac{n-1}{n(2n)} = \frac{n-1}{2n^2} $$ If the sum converged, we would have $$ \sum_{n=1}^\infty \frac{n-1}{2n^2} = \sum_{n=1}^\infty \frac{n}{2n^2} - \sum_{n=1}^\infty \frac{1}{2n^2} $$

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HINT: Observe that $$\sum_{n=1}^\infty\frac{n-1}{n(2n-1)}\ge\sum_{n=1}^\infty\frac{n-1}{n(2n)}=\sum_{n=1}^\infty\frac{1}{2n}-\sum_{n=1}^\infty\frac{1}{2n^2}.$$

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The question is not meaningful: The difference of the series is not defined since one (actually, each one) of them is not defined.

If, nevertheless, you insist on going ahead and "subtracting" them, then why not try it in the following artistic way? $$\sum_{n=1}^\infty \frac{1}{n}-\sum_{n=1}^\infty \frac{1}{2n-1}=$$$$\frac11-\left(\frac11\right)+\frac12-\left(\frac13+\frac15\right)+\frac13-\left(\frac17+\frac19+\frac1{11}\right)+ \frac14-\left(\frac1{13}+\frac1{15}+\frac1{17}+\frac1{19}\right)+\cdots,$$or, more formally,$$\sum_{n=1}^\infty \frac{1}{n}-\sum_{n=1}^\infty \frac{1}{2n-1}=\sum_{n=1}^\infty\left(\frac{1}{n}-\sum_{k=(n^2-n)/2+1}^{(n^2+n)/2} \frac{1}{2k-1}\right).$$The series on the right-hand side is well defined. It converges quite nicely to a smallish negative quantity.

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  • $\begingroup$ I can't decide whether you are serious or not. But in any case this answer is guaranteed to mislead the OP. $\endgroup$
    – TonyK
    Apr 7, 2017 at 12:36
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    $\begingroup$ The question is meaningful. The term "series" is not used strictly to refer to the sum, but also to the formal concept of a sequence of terms which is considered to be summed. The difference of two such series is the series formed from the term-by-term differences. This does not allow your rearrangement. $\endgroup$ Apr 7, 2017 at 15:13

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