sequences-and-series
Proving convergence:
$$\sum_{n=1}^\infty (-1)^{n-1}\frac1n$$
Just wanted to confirm if the reason they converge is due to the fact that for n= 1, 3, 5, … we have a positive harmonic series and for n= 2, 4, 6, … we have a negative harmonic series? therefore the two cancel giving a convergent sum.
Does the above proof make mathematical sense?
Is it possible for the difference of two particular (not the same) harmonic series to be divergent?
Here is a dataexample using the Euler-summation of negative instead of positive orders. I used the slowly divergent series $1+1/2+1/3+1/4+...$ and the sequences of partial sums using Eulersummation $ES(0)$ (direkt summation= no transformation) , Eulersummation $ES(-0.5)$ which has negative order and should accelerate divergence and Eulersummation $ES(-0.9)$ which accelerates divergence but even "overtunes" it: it makes it look like an alternating series. (All computations based on 32 elements) :
ES(0)(direct) ES(-0.5) ES(-0.9)
1.00000000000 1.00000000000 1.00000000000
1.50000000000 2.00000000000 6.00000000000
1.83333333333 2.33333333333 -5.66666666667
2.08333333333 2.66666666667 49.3333333333
2.28333333333 2.86666666667 -245.666666667
2.45000000000 3.06666666667 1526.00000000
2.59285714286 3.20952380952 -9861.85714286
2.71785714286 3.35238095238 67007.4285714
2.82896825397 3.46349206349 -471076.460317
2.92896825397 3.57460317460 3403128.53968
3.01987734488 3.66551226551 -25125107.3694
3.10321067821 3.75642135642 188836662.782
3.18013375513 3.83334443334 -1440564509.14
3.25156232656 3.91026751027 11129101675.0
3.31822899323 3.97693417693 -86914294560.5
3.38072899323 4.04360084360 685177450795.
3.43955252264 4.10242437301 -5.44613935055E12
3.49510807820 4.16124790242 4.36043950602E13
3.54773965714 4.21387948137 -3.51381487300E14
3.59773965714 4.26651106032 2.84800415982E15
3.64535870476 4.31413010794 -2.32041361096E16
3.69081325022 4.36174915556 1.89949738822E17
3.73429151109 4.40522741643 -1.56161905953E18
3.77595817775 4.44870567730 1.28888235269E19
3.81595817775 4.48870567730 -1.06760841088E20
3.85441971622 4.52870567730 8.87251757254E20
3.89145675325 4.56574271433 -7.39618656227E21
3.92717103897 4.60277975137 6.18296908223E22
3.96165379759 4.63726250999 -5.18235419676E23
3.99498713092 4.67174526861 4.35431150851E24
4.02724519544 4.70400333313 -3.66693900482E25
4.05849519544 4.73626139764 3.09468091836E26
However, I do not really think that this can be a general useful tool: since also convergent series might look like divergent by such "inverse" transformations. Here I used $1+1/2^2+1/3^2+1/4^2+...$
ES(0)(direct) ES(-0.5) ES(-0.9)
1.00000000000 1.00000000000 1.00000000000
1.25000000000 1.50000000000 3.50000000000
1.36111111111 1.44444444444 -7.88888888889
1.42361111111 1.55555555556 57.1111111111
1.46361111111 1.52888888889 -352.888888889
1.49138888889 1.58000000000 2402.38888889
1.51179705215 1.56390022676 -16936.9478458
1.52742205215 1.59383219955 123212.235828
1.53976773117 1.58289745528 -917619.148652
1.54976773117 1.60275636180 6963787.31960
1.55803219398 1.59477262837 -53665499.9384
1.56497663842 1.60900149714 418884302.009
1.57089379818 1.60287884486 -3305102741.43
1.57599583900 1.61362133802 26320630606.8
1.58044028344 1.60875562173 -211296315925.
1.58434653344 1.61717979015 1.70819287210E12
1.58780674106 1.61320690798 -1.38954751976E13
1.59089316081 1.62000633266 1.13658899049E14
1.59366324391 1.61669272000 -9.34278213695E14
1.59616324391 1.62230655034 7.71398168189E15
1.59843081761 1.61949494420 -6.39481412903E16
1.60049693331 1.62421545194 5.32067131888E17
1.60238729248 1.62179578230 -4.44177875633E18
1.60412340359 1.62582540701 3.71945515959E19
1.60572340359 1.62371815228 -3.12340250305E20
1.60720269353 1.62720177203 2.62971858850E21
1.60857443564 1.62534794030 -2.21942324134E22
1.60984994585 1.62839210680 1.87735425152E23
1.61103900649 1.62674694346 -1.59133257136E24
1.61215011760 1.62943185453 1.35152568303E25
1.61319070033 1.62796074625 -1.14995824956E26
1.61416726283 1.63034797478 9.80133223927E26
With $ES(-0.5)$ we get a sequence which lingers around at least in the near of the known result - but who would prognose that this extrapolates actually to a certain limit. And the sequence of partial sums of the $ES(-0.9)$-acceleration looks undoubtedly divergent...
So I think with the standard tools for acceleration/divergent summation taken only in an obviously/naively inverted way we are not yet really succeeding/proceeding in the desired direction...
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.
Best Answer
The two do not cancel at all. Perhaps you want to read about Leibniz Alternating Series
About your last question: for any $\,k,m\in\Bbb R\,$ , the series
$$\sum_{n=1}^\infty\left(\frac{1}{kn}-\frac{1}{mn}\right)=\frac{m-k}{mk}\sum_{n=1}^\infty\frac{1}{n}\,\,\,\text{diverges}$$