I’m interested if there are more techniques to make a series diverge “faster” to show it diverges. Below are the specific tricks/theorems I know of to do this.

I recalled reading that the sum of inverse primes 1/2+1/3+1/5+… has its nth partial sum growing like loglogn, which made me think that it would be potentially easier to show divergence by taking the exponential of the sum to make it diverge faster, and this indeed can be made to work to obtain a very short proof:

e∑∞n=12/pn>∞∏n=1(1+2/pn)>∞∏n=1∞∑k=01/pkn=∞∑n=11/n=∞

Similarly taking the exponential of the harmonic series and using ex>1+x gives a telescoping infinite product whose nth partial product is n+1, which clearly diverges.

I also recall a result that a series of positive decreasing terms ∑nan diverges if the series ∑k2ka2k diverges, and this transformation can also make a series diverge “faster”, e.g. it makes the harmonic series look like 1+1+1+… so that the harmonic series clearly diverges, and it makes ∑∞n=21/nlogn look like the harmonic series hence divergent.

Or a known divergent series could be used to construct a slower growing divergent series that we can then show grows the same or more slowly than a given series we wish to show divergent. I recall one such result that says that if a series of positive terms ∑nan diverges, then ∑nan/sn also diverges where sn=∑ni=1ai. Putting all an=1 shows the harmonic series diverges, and putting an=1/n shows that ∑∞n=21/nlogn diverges assuming that we know that the harmonic series partial sums grow asymptotically like logn.

Anyway, I was wondering if there are other theorems or tricks out there that show that a series diverges by doing something to make the series diverge “faster” if the series diverges (but obviously also leaves a convergent series still convergent). Or some alternative ways to take a divergent series and construct a slower growing divergent series. E.g. in a comment someone pointed out “Series Acceleration” methods designed to make a convergent series converge faster; if someone can show an example of how these techniques can also be used to make an example divergent series diverge faster to show divergence, that would be great.

**Answer**

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+1/22+1/32+1/42+...

```
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…

**Attribution***Source : Link , Question Author : user2566092 , Answer Author : Gottfried Helms*