Is the series $\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots$ summable?

Question

Is the series $\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots$ summable? I think it diverges although:

$$ \sqrt{n+1} - \sqrt{n} \approx \frac{1}{2\sqrt{n}}$$ for example by the Mean Value Theorem $f(x+1)-f(x) \approx f'(x)$ and then I might argue: $$ \sum_{n \geq 1} (-1)^{n+1} \sqrt{n} = \frac{1}{2}\sum_{m \geq 1} \frac{1}{\sqrt{2m}} = \infty $$ Are these Cesaro summable? For an even number of terms: $$\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots - \sqrt{2n} \approx - \frac{1}{2\sqrt{2}}\left( \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \dots + \frac{1}{\sqrt{n}} \right) \approx \sqrt{\frac{n}{2}}$$ so the Cesaro means tend to infinity. Does any more creative summation method work?

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The result is from paper called "The Second Theorem of Consistency for Summable Series" in Vol 6 of the Collected Works of GH Hardy

the series $1 - 1 +1 - 1 \dots$ is summable $(1,k)$ for any $k$ but not summable $(e^n, k)$ for any value of $k$.

The series $\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots$ is summable $(n,1)$ but not $(e^{\sqrt{n}},1)$ and so on...

Here things like $(1,k), (n,1)$ refer to certain averaging procedures, IDK

Answer 1

Let us show that $\sum_{k=1}^{\infty}(-1)^{k-1}\sqrt{k}$ is Cesaro summable. Once we establish this, then this is also Abel summable and the Cesaro sum is equal the Abel sum, which is

$$ \sum_{k=1}^{\infty}(-1)^{k-1}\sqrt{k} = -\operatorname{Li}_{-1/2}(-1) = (1 - 2^{3/2})\zeta(-1/2). $$

To this end, let $s_n = \sum_{k=1}^{n} (-1)^{k-1}\sqrt{k}$ and notice that $s_n = \mathcal{O}(\sqrt{n})$. This can be easily checked by grouping two successive terms and applying the mean value theorem. Thus it suffices to prove that

$$ \frac{s_1 + \cdots + s_{2n+1}}{2n+1} $$

converges. Now the trick is to consider

$$ s_{2n} + s_{2n+1} = 1 + \sum_{k=1}^{n} (\sqrt{2k-1} + \sqrt{2k+1} - 2\sqrt{2k}). $$

Using Taylor series, it is not hard to check that

$$\sqrt{2k-1} + \sqrt{2k+1} - 2\sqrt{2k} = \mathcal{O}(k^{-3/2}). $$

Thus $s_{2n} + s_{2n+1}$ converges as $n \to \infty$, and the claim follows from Cearso-Stolz theorem.

Answer 2

How about an Abel sum? $$ \sum_{n=1}^\infty (-1)^{n+1}\sqrt{n}\;x^n = -\mathrm{Li}_{-1/2}(-x) $$ for $|x|<1$ and converges as $x \to 1^-$ to the value $$ -\mathrm{Li}_{-1/2}(-1) \approx 0.3801048 $$ So we call that value the Abel sum of the divergent series $\sum_{n=1}^\infty (-1)^{n+1}\sqrt{n}$

Answer 3

This sum can be done with some form of zeta function regularization. For $\Re s >1$, define:

$$\eta(s) = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} = (1-2^{1-s})\zeta(s) $$

Then, by analytic continuation, we can calculate:

$$\sum_{n=1}^\infty (-1)^{n-1} \sqrt{n} \to \eta\left(-\frac{1}{2}\right) = (1-2\sqrt{2})\zeta\left(-\frac{1}{2}\right) \approx .3801048$$

This is equal to the Abel sum $-\operatorname{Li}_{-\frac{1}{2}} (-1)$ from GEdgar's answer.

Answer 4

I thought it might be instructive to present a brute force approach. To that end we proceed.

Let $S_n=\sum_{k=1}^n (-1)^{k-1}\sqrt{k}$ be the sequence of interest. Then, we can write the even and odd terms, respectively by

$$\begin{align} S_{2n}&=\sum_{k=1}^n \left(\sqrt{2k-1}-\sqrt{2k}\right)\\\\ S_{2n+1}&=1+\sum_{k=1}^n \left(\sqrt{2k+1}-\sqrt{2k}\right) \end{align}$$

Then, the Cesaro Sum is given by

$$\begin{align} \frac{\sum_{n=0}^N (S_{2n}+S_{2n+1})}{2N+1}&=\frac{1+\sum_{n=1}^N\left(1+ \sum_{k=1}^n \left(\sqrt{2k+1}-2\sqrt{2k}+\sqrt{2k-1}\right)\right)}{2N+1} \tag 1\\\\ &=\frac{N+1}{2N+1}+\frac{1}{2N+1}\sum_{n=1}^N\sum_{k=1}^n \left(\sqrt{2k+1}-2\sqrt{2k}+\sqrt{2k-1}\right)\tag 2\\\\ &=\frac{N+1}{2N+1}+\frac{1}{2N+1}\sum_{k=1}^N\sum_{n=k}^N \left(\sqrt{2k+1}-2\sqrt{2k}+\sqrt{2k-1}\right)\tag 3\\\\ &=\frac{N+1}{2N+1}+\frac{1}{2N+1}\sum_{k=1}^N(N+1-k)\left(\sqrt{2k+1}-2\sqrt{2k}+\sqrt{2k-1}\right) \tag 4\\\\ &=\frac{N+1}{2N+1}+\frac{N+1}{2N+1}\sum_{k=1}^N \left(\sqrt{2k+1}-2\sqrt{2k}+\sqrt{2k-1}\right) \tag 5\\\\ &-\frac{1}{2N+1}\sum_{k=1}^N k\left(\sqrt{2k+1}-2\sqrt{2k}+\sqrt{2k-1}\right) \end{align}$$

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NOTES:

In going from $(1)$ to $(2)$, we simply carried our the trivial sum $\sum_{k=1}^n (1)=N$.

In going from $(2)$ to $(3)$, we interchanged the order of summation.

In going from $(3)$ to $(4)$, we evaluated the inner sum.

In going from $(4)$ to $(5)$, we split the expression into the sum of three terms.

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As $N\to \infty$, the first term in $(5)$ approaches $\frac12$ while the second term approaches $\frac12 \sum_{k=1}^\infty \left(\sqrt{2k+1}-2\sqrt{2k}+\sqrt{2k-1}\right)$. The third term in $(5)$ can be shown to approach $0$ since $\sum_{k=1}^N k \left(\sqrt{2k+1}-2\sqrt{2k}+\sqrt{2k-1}\right) =O\left(N^{1/2}\right)$.

Therefore, we find that

$$\lim_{N\to \infty}\frac{\sum_{n=0}^N (S_{2n}+S_{2n+1})}{2N+1}=\frac12+\frac12 \sum_{k=1}^\infty \left(\sqrt{2k+1}-2\sqrt{2k}+\sqrt{2k-1}\right) $$

Answer 5

For the intuition - here are some summations, showing the progress of the partial sums up to index n [update] I inserted the comparison with my home-brewed "Eulerian"-summation, which is related to the Borel-summation):

   n    direct sum      Cesarosum       Cesarosum       Eulersum         Eulersum        Eulersum        "Eulerian"sum
                          order 2         order 4         order 1         order 1/2     order sqrt(1/2)
   -------------------------------------------------------------------------------------------------------------------
   0    1.00000000000   1.00000000000   1.00000000000   1.00000000000    1.00000000000   1.00000000000  0.194415575888
   1  -0.414213562373  0.292893218813  0.646446609407  0.292893218813  0.0571909584179  0.171572875254  0.378140142714
   2    1.31783724520  0.634541227608  0.607519655808  0.372352530112   0.512721636810  0.422772999027  0.389288545829
   3  -0.682162754804  0.305365232005  0.539303380327  0.378588536708   0.328572723321  0.370989365477  0.379768169295
   4    1.55390522270  0.555073230143  0.522065851592  0.379713964073   0.399353836977  0.381763811860  0.379611976106
   5  -0.895584520088  0.313296938438  0.493541493492  0.379988083920   0.373053691245  0.379763647612  0.380149100952
   6    1.75016679098  0.518564060229  0.483741558178  0.380066565914   0.382646545140  0.380160944951  0.380128967709
   7   -1.07826033377  0.318961010979  0.468182885218  0.380091452109   0.379197665397  0.380092326363  0.380101202755
   8    1.92173966623  0.497047528229  0.461836541330  0.380099920721   0.380425772282  0.380106436474  0.380103740014
   9   -1.24053799394  0.323288976013  0.452077282604  0.380102955702   0.379991936204  0.380104311828  0.380105054843
  10    2.07608679642  0.482634232413  0.447622208332  0.380104087332   0.380144293995  0.380104835957  0.380104854942
  11   -1.38801481872  0.326746811485  0.440944677832  0.380104522684   0.380091058757  0.380104786215  0.380104798120
  12    2.21753645674  0.472192168813  0.437640158402  0.380104694466   0.380109586124  0.380104809822  0.380104811220
  13   -1.52412093003  0.329598376039  0.432791176214  0.380104763682   0.380103160679  0.380104810527  0.380104813408
  14    2.34886241618  0.464215978714  0.430239830835  0.380104792065   0.380105382732  0.380104812081  0.380104812638
  15   -1.65113758382  0.332006381056  0.426562761461  0.380104803879   0.380104616275  0.380104812387  0.380104812569
  16    2.47196804180  0.457886478746  0.424531900800  0.380104808861   0.380104880083  0.380104812536  0.380104812610
  17   -1.77067264532  0.334077638520  0.421650006923  0.380104810985   0.380104789461  0.380104812582  0.380104812612
  18    2.58822629822  0.452717041662  0.419994156166  0.380104811900   0.380104820539  0.380104812600  0.380104812610
  19   -1.88390965678  0.335885706740  0.417676078989  0.380104812297   0.380104809898  0.380104812606  0.380104812610
  20    2.69866603817  0.448399055856  0.416299534590  0.380104812471   0.380104813536  0.380104812608  0.380104812610
  21   -1.99174972165  0.337483202333  0.414395464588  0.380104812548   0.380104812294  0.380104812609  0.380104812610
  22    2.80408180166  0.444726619695  0.413232642238  0.380104812582   0.380104812717  0.380104812610  0.380104812610
  23   -2.09489768390  0.338908940378  0.411641364821  0.380104812597   0.380104812573  0.380104812610  0.380104812610
  24    2.90510231610  0.441556675407  0.410645790363  0.380104812604   0.380104812622  0.380104812610  0.380104812610
  25   -2.19391719750  0.340192295680  0.409296497183  0.380104812607   0.380104812605  0.380104812610  0.380104812610
  26    3.00223522521  0.438786478255  0.408434304228  0.380104812609   0.380104812611  0.380104812610  0.380104812610
  27   -2.28926739692  0.341355982713  0.407275994591  0.380104812609   0.380104812609  0.380104812610  0.380104812610
  28    3.09589741021  0.436340169868  0.406521911759  0.380104812609   0.380104812610  0.380104812610  0.380104812610
  29   -2.38132816484  0.342417892045  0.405516932547  0.380104812610   0.380104812610  0.380104812610  0.380104812610
  30    3.18643619799  0.434160418043  0.404851713796  0.380104812610   0.380104812610  0.380104812610  0.380104812610
  31   -2.47041805150  0.343392340870  0.403971669522  0.380104812610   0.380104812610  0.380104812610  0.380104812610

Cesarosum(2) seems to be too weak; it gives with 32 partial sums still oscillating values around the final value $s$. But the interval appears to be contracting so it seems it shall be sufficient in principle. Cesarosum(4) does the job better, and while approximating monotonously from above it does no more oscillations.
Cesarosums can also be generalized to fractional orders, but I didn't try to find the optimal fractional order at the moment.

Euler-summation can sum the alternating harmonic series of any exponent (this can be seen by G.H.Hardy's or by K.Knopp's books); here Eulersum(1) gives a standard solution, but may be seen to approximate "too slow" (so we may say it is "too strong" for this series) while Eulersum($\frac12$) is "too weak" - at 26'th term the partial sums are still visibly oscillating around $s$ and may continue so in trailing (non displayed) digits; however the interval is sharply contracting, so it shall find better approximations later. Eulersum($\sqrt{\frac12}$) seems to be better for this case of series because the correct first 12 digits occur as constant result earliest in that little comparision.
A self-invented "Eulerian"-summation (as I called it) converges even quicker. The Eulerian-summation can sum alternating harmonic/geometric/hypergeometric (factorials) series, and looks a bit like Borel-summation.

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To see a bit more, here are the differences of the partial sums to the final value:

   n   direct sum     Cesarosum    Cesarosum           Eulersum               Eulersum             Eulersum        "Eulerian"sum
                        order 2      order 4            order 1               order 1/2       order sqrt(1/2)
   --------------------------------------------------------------------------------------------------------------------------------
   0   0.61989519    0.61989519   0.61989519            0.61989519            0.61989519            0.61989519          -0.18568924
   1  -0.79431837  -0.087211594   0.26634180          -0.087211594           -0.32291385           -0.20853194        -0.0019646699
   2   0.93773243    0.25443641   0.22741484         -0.0077522825            0.13261682           0.042668186         0.0091837332
   3   -1.0622676  -0.074739581   0.15919857         -0.0015162759          -0.051532089         -0.0091154471       -0.00033664331
   4    1.1738004    0.17496842   0.14196104        -0.00039084854           0.019249024          0.0016589993       -0.00049283650
   5   -1.2756893  -0.066807874   0.11343668        -0.00011672869         -0.0070511214        -0.00034116500       0.000044288342
   6    1.3700620    0.13845925   0.10363675       -0.000038246696          0.0025417325        0.000056132342       0.000024155100
   7   -1.4583651  -0.061143802  0.088078073       -0.000013360501        -0.00090714721       -0.000012486246     -0.0000036098548
   8    1.5416349    0.11694272  0.081731729      -0.0000048918883         0.00032095967       0.0000016238648     -0.0000010725957
   9   -1.6206428  -0.056815837  0.071972470      -0.0000018569080        -0.00011287641     -0.00000050078187     0.00000024223295
  10    1.6959820    0.10252942  0.067517396     -0.00000072527779        0.000039481386     0.000000023346826    0.000000042331835
  11   -1.7681196  -0.053358001  0.060839865     -0.00000028992563       -0.000013753852    -0.000000026395105   -0.000000014489245
  12    1.8374316   0.092087356  0.057535346     -0.00000011814347       0.0000047735140   -0.0000000027880405  -0.0000000013892189
  13   -1.9042257  -0.050506437  0.052686364    -0.000000048928020      -0.0000016519303   -0.0000000020824686  0.00000000079841255
  14    1.9687576   0.084111166  0.050135018    -0.000000020544902      0.00000057012251  -0.00000000052824026        2.8413518E-11
  15   -2.0312424  -0.048098432  0.046457949   -0.0000000087303762     -0.00000019633449        -2.2297034E-10       -4.1154003E-11
  16    2.0918632   0.077781666  0.044427088   -0.0000000037487327     0.000000067473071        -7.3301616E-11        6.9589588E-13
  17   -2.1507775  -0.046027174  0.041545194   -0.0000000016244918    -0.000000023148630        -2.7573817E-11        1.9964371E-12
  18    2.2081215   0.072612229  0.039889344  -0.00000000070971290    0.0000000079289801        -9.8174144E-12       -1.3600703E-13
  19   -2.2640145  -0.044219106  0.037571266  -0.00000000031232210   -0.0000000027121709        -3.6279833E-12       -9.1097422E-14
  20    2.3185612   0.068294243  0.036194722        -1.3834315E-10   0.00000000092651001        -1.3293701E-12        1.1293573E-14
  21   -2.3718545  -0.042621610  0.034290652        -6.1641605E-11  -0.00000000031615339        -4.9330835E-13        3.8785395E-15
  22    2.4239770   0.064621807  0.033127830        -2.7613136E-11         1.0776531E-10        -1.8338260E-13       -7.4767696E-16
  23   -2.4750025  -0.041195872  0.031536552        -1.2430206E-11        -3.6699083E-11        -6.8598280E-14       -1.5092194E-16
  24    2.5249975   0.061451863  0.030540978        -5.6206028E-12         1.2486495E-11        -2.5751761E-14        4.3987037E-17
  25   -2.5740220  -0.039912517  0.029191685        -2.5519505E-12        -4.2450512E-12        -9.7087865E-15        5.0971977E-18
  26    2.6221304   0.058681666  0.028329492        -1.1630745E-12         1.4420926E-12        -3.6731229E-15       -2.3894899E-18
  27   -2.6693722  -0.038748830  0.027171182        -5.3194349E-13        -4.8956357E-13        -1.3944552E-15       -1.2586396E-19
  28    2.7157926   0.056235357  0.026417099        -2.4408222E-13         1.6608839E-13        -5.3103197E-16        1.2185701E-19
  29   -2.7614330  -0.037686921  0.025412120        -1.1233653E-13        -5.6313726E-14        -2.0282063E-16       -7.7665569E-23
  30    2.8063314   0.054055605  0.024746901        -5.1847946E-14         1.9082742E-14        -7.7675619E-17       -5.8729086E-21
  31   -2.8505229  -0.036712472  0.023866857        -2.3993181E-14        -6.4631442E-15        -2.9824069E-17        3.0005419E-22

I think it is better visible now, that the Cesaro-summations approximate far slower than all the Euler-summations; also showing alternating signs in the distances from the final value $s$ may be seen as useful as long as the intervals contract (and even contract faster than that of the non-alternating distances).

[update] Just for completion: the Cesarosum(3) seems to work better than Cesarosum(4), and even better the Cesarosum of fractional order 2.8. Whether for instance Cesarosum(2.35) is even better is a matter of opinion, though... See the approximation-errors of the partial sums around n=32-terms:

   showing (n'th Partial sums) - (final value)
  ------------------------------------------------------------------------
   n !  Cesarosum(4)  ! Cesarosum(3)  ! Cesarosum(2.8)   ! Cesarosum(2.35)
  ---+----------------+---------------+------------------+-----------------
   ...       ...          ...                 ...
  25  0.0291916845736  0.0179475747469  0.0144055813350   -0.00148866655773
  26  0.0283294916179  0.0208571526683  0.0205452310850     0.0275184163551
  27  0.0271711819811  0.0167463952500  0.0134610730999   -0.00152042853712
  28  0.0264170991491  0.0193789927172  0.0190753425386     0.0257585187486
  29  0.0254121199376  0.0156973208926  0.0126350050556   -0.00154129743087
  30  0.0247469011866  0.0180947136764  0.0177989024790     0.0242186544925
  31  0.0238668569121  0.0147730660754  0.0119062541335   -0.00155396729937
   ...       ...          ...                 ...

Answer 6

Hopefully I'm not copying Gottfried Helms' Euler sum, but the Euler summation formula allows you to sum this divergent series.

$$\sum_{k=1}^\infty(-1)^{k+1}\sqrt k=\sum_{n=0}^\infty\frac1{2^n}\sum_{k=0}^n\binom nk(-1)^k\sqrt{k+1}$$

This converges extremely quickly to the Riemann zeta function,

$$\zeta(-1/2)=\frac1{1-\sqrt8}\sum_{n=0}^\infty\frac1{2^n}\sum_{k=0}^n\binom nk(-1)^k\sqrt{k+1}$$