86
$\begingroup$

The following is probably a math contest problem. I have been unable to locate the original source.

Suppose that $\{a_i\}$ is a set of positive real numbers and the series $\sum_{n = 1}^\infty \frac{1}{a_n}$ converges.

Show $\sum_{n = 1}^\infty \frac{n^2a_n}{(a_1+\cdots+a_n)^2}$

also converges.

  • 0
    Thanks for finding the source!2012-09-29

6 Answers 6

47

Define at first some quantities to simplify the typing for the rest of the proof

  • $C^2:=\sum_{n=1}^{+\infty}\frac{1}{a_n}.$
  • $A_n=a_1+\dotso+a_n.$

Moreover let $P_N:=\sum_{n=1}^N\frac{n^2a_n}{(a_1+\dotso+a_n)^2}.$ Clearly $P_{N+1}>P_N$, that is, $P_N$ is an increasing sequence. If we can prove that it is also bounded above, we are done with the proof. To reach this goal, notice that $\begin{split}P_N<&\frac{1}{a_1}+\sum_{n=2}^N\frac{n^2(A_n-A_{n-1})}{A_nA_{n-1}}\\ =&\frac{1}{a_1}+\sum_{n=2}^N\left(\frac{n^2}{A_{n-1}}-\frac{n^2}{A_n}\right).\end{split}\tag{1}$ Since $(n+1)^2-n^2=2n+1<5n$ for every $n\in\mathbb N$, one gets from $(1)$ that $\begin{split}P_N<&\frac{1}{a_1}+\frac{4}{a_1}+\sum_{n=2}^{N-1}\frac{2n+1}{A_n}-\frac{N^2}{A_N}\\ <&\frac{5}{a_1}+\frac{5}{A_2}+\dots+\frac{2N-1}{A_{N-1}}-\frac{N^2}{A_N}\\<&5\left(\frac{1}{A_1}+\frac{2}{A_2}+\dots+\frac{N}{A_N}\right).\end{split}\tag{2}$ By Cauchy Schwarz we also have $\sqrt{\left(\frac{1}{a_1}+\dots+\frac{1}{a_N}\right)}\sqrt{\left(\frac{a_1}{A_1^2}+\dots+\frac{N^2a_N}{A_N^2}\right)}\geq\left(\frac{1}{A_1}+\frac{2}{A_2}+\dots+\frac{N}{A_N}\right),\tag{3}$ from which, following $(2)$, it turns out that $P_N<5C\sqrt{P_N}.$ It is then clear that the sequence $P_N$ is bounded from above, since for any $N\in\mathbb N$, we have estabilished $P_N<25C^2.$ Therefore, since $P_N$ is also increasing as observed at the beginning, we can conclude that $P_N$ converges. This concludes the proof.

  • 0
    Simply amazing !!!! :QD2017-01-28
9

I wrote this answer for the closed duplicate of this question, but it works here as well.

Define $ \bar{p}_n=\frac1n\sum_{k=1}^np_k\tag{1} $ then the series in question is $ \sum_{k=1}^\infty\frac{p_k}{\bar{p}_k^2}\tag{2} $ Simply, for $n\ge m$, we have that $ \bar{p}_n=\frac1n\sum_{k=1}^np_k\ge\frac mn\frac1m\sum_{k=1}^mp_k=\frac mn\bar{p}_m\tag{3} $ which, for $n\ge1$, says that $ \bar{p}_{n+1}\ge\frac12\bar{p}_n\quad\text{and}\quad\bar{p}_{2n+1}\ge\frac23\bar{p}_{2n}\tag{4} $ Furthermore, $ \begin{align} \sum_{k=1}^\infty\frac{p_{k+1}}{\bar{p}_{k+1}\bar{p}_k} &=\sum_{k=1}^\infty\frac{(k+1)\bar{p}_{k+1}-k\bar{p}_k}{\bar{p}_{k+1}\bar{p}_k}\\ &=\sum_{k=1}^\infty\left(\frac{k}{\bar{p}_k}+\frac1{\bar{p}_k}-\frac{k+1}{\bar{p}_{k+1}}+\frac1{\bar{p}_{k+1}}\right)\\ &=2\sum_{k=1}^\infty\frac1{\bar{p}_k}\tag{5} \end{align} $ Combining $(4)$ and $(5)$ yields $ \begin{align} \sum_{k=1}^\infty\frac{p_k}{\bar{p}_k^2} &=\frac1{p_1}+\sum_{k=1}^\infty\frac{p_{k+1}}{\bar{p}_{k+1}^2}\\ &\le\frac1{p_1}+2\sum_{k=1}^\infty\frac{p_{k+1}}{\bar{p}_{k+1}\bar{p}_k}\\ &=\frac1{p_1}+4\sum_{k=1}^\infty\frac1{\bar{p}_k}\tag{6} \end{align} $ Use $\color{#C00000}{(4)}$, $\color{#00A000}{\text{Jensen's Inequality}}$, and change the $\color{#0000FF}{\text{order of summation}}$ to get $ \begin{align} \sum_{k=1}^\infty\frac1{\bar{p}_k} &=\frac1{p_1}+\sum_{k=1}^\infty\left(\frac1{\bar{p}_{2k}}+\frac1{\bar{p}_{2k+1}}\right)\\ &\le\frac1{p_1}+\color{#C00000}{\frac52\sum_{k=1}^\infty\frac1{\bar{p}_{2k}}}\\ &\le\frac1{p_1}+5\sum_{k=1}^\infty\frac1{\displaystyle\small\frac2{2k}\sum_{j=k+1}^{2k}p_j}\\ &\le\frac1{p_1}+5\sum_{k=1}^\infty\color{#00A000}{\frac1k\sum_{j=k+1}^{2k}\frac1{p_j}}\\ &=\frac1{p_1}+5\color{#0000FF}{\sum_{j=2}^\infty\frac1{p_j}\sum_{k=\lceil j/2\rceil}^{j-1}\frac1k}\\ &\le\frac1{p_1}+5\sum_{j=2}^\infty\frac1{p_j}\tag{7} \end{align} $ Combining $(6)$ and $(7)$ gives $ \sum_{k=1}^\infty\frac{p_k}{\bar{p}_k^2}\le20\sum_{j=1}^\infty\frac1{p_j}\tag{8} $

3

I printed out the Purdue problem. I also felt it worthwhile to do some experiments. If we stick with $a_n = n^k$ for fixed $k \geq 2,$ things are pretty good. So what I did was run $ a_n = (n + 14) \log(n+14) \left( \log \log (n+14) \right)^2 \; \; / \; \; 15. $ Note that, in contrast to the Purdue problem it is possible for the ratio to exceed $e.$ Also, the ratio is steadily increasing, with the appearance of approaching a limit. I have the feeling that if I changed the "fancy" sum to have $(n + 14)^2$ instead of $n^2$ in the numerator, the ratio might easily stay below $e.$ I should try that. NO: with the $(n + 14)^2$ the ratio starts at 225 and decreases. Annoying, but better to know the truth.

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    n   harmonic sum           fancy sum             ratio     1   0.3720703157450488    0.3720703157450488      1     2   0.697211225420433    0.7424549077736787      1.064892360741844     3   0.9843674932090888    1.109100676396848      1.126714041298867     4   1.240296228954154    1.47046914150556      1.185578982809206     5   1.470203813539751    1.825431331069707      1.241617872473538     6   1.678171381303085    2.173181689415959      1.29497005706801     7   1.867446704822137    2.513168481471731      1.345777887520006     8   2.040649149703482    2.845038024641238      1.394182838855292     9   2.199916583903517    3.168590269249502      1.440322916074925    10   2.347012629961203    3.483743567272218      1.484330984332967    11   2.483406263112077    3.790506814518016      1.526333758121375    12   2.610331767584863    4.088957475567311      1.566451255868715    13   2.728834508384516    4.379224284501249      1.604796579288999    14   2.839806304031565    4.661473653263476      1.641475915679798    15   2.94401306997052    4.935899015735965      1.676588689800005    16   3.042116644209738    5.202712494418531      1.71022781270442    17   3.134692183052954    5.462138403848396      1.742479990022078    18   3.222242147523305    5.714408206199048      1.773426063150246    19   3.305207639920737    5.959756614818636      1.803141364807435    20   3.383977661836249    6.198418604998871      1.831696076160095    21   3.458896727815719    6.430627141452748      1.859155576903756    22   3.530271167787264    6.656611471609346      1.885580782674448    23   3.598374376086184    6.876595865132866      1.911028466307689    24   3.663451208300954    7.090798704807332      1.935551560981734    25   3.72572168420513    7.299431853492455      1.959199444348663    26   3.785384122162883    7.502700237347626      1.982018203494961    27   3.842617805032313    7.700801597802712      2.004050881073236    28   3.897585257875515    7.893926374503574      2.025337703274867    29   3.950434202350573    8.082257689209232      2.045916290518181    30   4.001299240496784    8.26597140678431      2.065821851843813    31   4.050303310976641    8.445236254343797      2.085087364063956    32   4.097558953139548    8.620213983526074      2.103743736724342    33   4.143169408093855    8.791059563998846      2.12181996392065    34   4.187229580988316    8.957921398802352      2.139343263974555    35   4.229826884660201    9.120941554132628      2.156339207926083    36   4.271041981511019    9.280255997765739      2.172831837743385    37   4.310949437771172    9.43599484160206      2.188843775092063    38   4.349618302093892    9.5882825848318      2.204396321446421    39   4.387112618583768    9.737238355039727      2.219509550266131    40   4.423491882842733    9.882976145219228      2.234202391904919    41   4.458811448348255    10.02560504518575      2.24849271186376    42   4.493122889418268    10.16522946629327      2.262397382950141    43   4.526474326127627    10.30194935868578      2.275932351857398    44   4.558910715791747    10.43586042057516      2.289112700633962    45   4.590474115000052    10.56705429924105      2.301952703471661    46   4.62120391564534    10.6956187836084      2.314465879204723    47   4.651137057938973    10.82163798838195      2.326665039876781    48   4.680308223012721    10.94519252981278      2.338562335701759    49   4.708750007375382    11.06635969324407      2.350169296715833    50   4.73649308120685    11.1852135926374      2.361496871391487    51   4.763566332226935    11.30182532231984      2.372555462460898    52   4.789996996664597    11.41626310121996      2.383354960174174    53   4.815810778670352    11.52859240987866      2.393904773198276    54   4.841031959356151    11.63887612053186      2.404213857344543    55   4.865683496509348    11.74717462056706      2.414290742296434    56   4.889787115907601    11.85354592965637      2.424143556494324    57   4.913363395057002    11.95804581086575      2.433780050320707    58   4.93643184008435    12.06072787603474      2.443207617716981    59   4.959010956434438    12.16164368571353      2.452433316351822    60   4.981118313952936    12.26084284393527      2.461463886451085    61   5.002770606873656    12.35837308809172      2.470305768389958    62   5.023983709174518    12.45428037416987      2.478965119139729    63   5.044772725718442    12.54860895759598      2.487447827653899    64   5.06515203955289    12.64140146992287      2.495759529271456    65   5.085135355704092    12.73269899158463      2.503905619208763    66   5.104735741768568    12.82254112093224      2.511891265205786    67   5.123965665574847    12.91096603975298      2.519721419387091    68   5.142837030161822    12.99801057546534      2.527400829393256    69   5.161361206296602    13.08371026017186      2.534934048833937    70   5.179549062733651    13.16809938674168      2.542325447109839    71   5.197410994398156    13.25121106208583      2.549579218647163    72   5.214956948659732    13.33307725777886      2.556699391584722    73   5.232196449847382    13.41372885817208      2.563689835950893    74   5.2491386221431    13.49319570613536      2.5705542713647    75   5.265792210979244    13.57150664655682      2.577296274292794    76   5.282165603053812    13.64868956772222      2.583919284891677    77   5.29826684506785    13.72477144068912      2.59042661346238    78   5.314103661280238    13.79977835676407      2.596821446542787    79   5.329683469967026    13.87373556318502      2.603106852660962    80   5.345013398865164    13.9466674971053      2.609285787771161    81   5.360100299673816    14.01859781796976      2.615361100392628    82   5.374950761680474    14.08954943836886      2.621335536469878    83   5.389571124573571    14.15954455345108      2.627211743971818    84   5.403967490498364    14.22860466896988      2.632992277245858    85   5.418145735408332    14.29675062803671      2.638679601142042    86   5.432111519760208    14.36400263664786      2.644276094921177    87   5.445870298597022    14.43038028804854      2.649784055960005    88   5.459427331060111    14.49590258599466      2.655205703265556    89   5.472787689367907    14.56058796696874      2.660543180810006    90   5.485956267296481    14.62445432140357      2.665798560696622    91   5.498937788194169    14.68751901396411      2.67097384616665    92   5.511736812560254    14.74979890293522      2.67607097445638    93   5.524357745215423    14.81131035876027      2.681091819512984    94   5.536804842089758    14.87206928177309      2.686038194577204    95   5.549082216652123    14.9320911191635      2.690911854640413    96   5.561193846003126    14.99139088121412      2.695714498783125    97   5.573143576652275    15.04998315684457      2.700447772401537    98   5.584935129998476    15.10788212849676      2.70511326932832    99   5.59657210753173    15.16510158639337      2.709712533853454   100   5.60805799577263    15.22165494219984      2.714247062650558   101   5.61939617096513    15.2775552421185      2.718718306613819   102   5.630589903537044    15.332815179442      2.723127672610321   103   5.641642362341722    15.38744710659176      2.72747652515229   104   5.652556618693491    15.44146304666565      2.731766187993482   105   5.663335650208609    15.49487470451806      2.735997945653691   106   5.673982344462718    15.54769347739415      2.740173044875133   107   5.684499502475065    15.59993046513886      2.744292696014233   108   5.694889842029118    15.6515964800005      2.74835807437212   109   5.705156000838572    15.70270205604725      2.752370321466966   110   5.715300539567187    15.75325745821449      2.756330546251111   111   5.725325944710379    15.80327269099944      2.760239826275753   112   5.735234631345977    15.85275750681911      2.764099208805813   113   5.745028945761108    15.9017214140467      2.767909711887452   114   5.754711167961762    15.95017368474058      2.771672325370572   115   5.764283514071177    15.99812336207967      2.775388011888501 

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  • 0
    @alex.jordan, that sounds interesting. Not sure what to expect.2012-09-25
2

As suggested by Alex Jordan, here is the same program with $ a_n = n! $ We know that the sum is $e-1.$ I'm not sure what the fancy sum is, but it also converges rapidly. I stopped this at $n=69$ because my calculator says $70! > 10^{100}.$

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jagy@phobeusjunior:~$      jagy@phobeusjunior:~$ ./series_potato      n    harmonic             fancy                  ratio     1   1                    1                      1     2   1.5                  1.888888888888889      1.259259259259259     3   1.666666666666667    2.555555555555555      1.533333333333333     4   1.708333333333333    2.908172635445363      1.702344957333871     5   1.716666666666667    3.036328472943761      1.768735032782773     6   1.718055555555556    3.070338463378451      1.78710080325989     7   1.718253968253968    3.077401815520791      1.791005213651823     8   1.71827876984127    3.07860906311077      1.791681953560518     9   1.718281525573192    3.078784678063843      1.791781283941121    10   1.718281801146385    3.078806934166605      1.791793949113888    11   1.718281826198493    3.07880943411009      1.79179537789887    12   1.718281828286169    3.078809686323769      1.791795522504363    13   1.718281828446759    3.078809709421133      1.79179553577903    14   1.71828182845823    3.078809711357875      1.791795536894207    15   1.718281828458995    3.078809711507647      1.791795536980574    16   1.718281828459042    3.078809711518395      1.791795536986779    17   1.718281828459045    3.078809711519114      1.791795536987195    18   1.718281828459046    3.078809711519159      1.791795536987221    19   1.718281828459046    3.078809711519162      1.791795536987222    20   1.718281828459046    3.078809711519162      1.791795536987222    21   1.718281828459046    3.078809711519162      1.791795536987222    22   1.718281828459046    3.078809711519162      1.791795536987222    23   1.718281828459046    3.078809711519162      1.791795536987222    24   1.718281828459046    3.078809711519162      1.791795536987222    25   1.718281828459046    3.078809711519162      1.791795536987222    26   1.718281828459046    3.078809711519162      1.791795536987222    27   1.718281828459046    3.078809711519162      1.791795536987222    28   1.718281828459046    3.078809711519162      1.791795536987222    29   1.718281828459046    3.078809711519162      1.791795536987222    30   1.718281828459046    3.078809711519162      1.791795536987222    31   1.718281828459046    3.078809711519162      1.791795536987222    32   1.718281828459046    3.078809711519162      1.791795536987222    33   1.718281828459046    3.078809711519162      1.791795536987222    34   1.718281828459046    3.078809711519162      1.791795536987222    35   1.718281828459046    3.078809711519162      1.791795536987222    36   1.718281828459046    3.078809711519162      1.791795536987222    37   1.718281828459046    3.078809711519162      1.791795536987222    38   1.718281828459046    3.078809711519162      1.791795536987222    39   1.718281828459046    3.078809711519162      1.791795536987222    40   1.718281828459046    3.078809711519162      1.791795536987222    41   1.718281828459046    3.078809711519162      1.791795536987222    42   1.718281828459046    3.078809711519162      1.791795536987222    43   1.718281828459046    3.078809711519162      1.791795536987222    44   1.718281828459046    3.078809711519162      1.791795536987222    45   1.718281828459046    3.078809711519162      1.791795536987222    46   1.718281828459046    3.078809711519162      1.791795536987222    47   1.718281828459046    3.078809711519162      1.791795536987222    48   1.718281828459046    3.078809711519162      1.791795536987222    49   1.718281828459046    3.078809711519162      1.791795536987222    50   1.718281828459046    3.078809711519162      1.791795536987222    51   1.718281828459046    3.078809711519162      1.791795536987222    52   1.718281828459046    3.078809711519162      1.791795536987222    53   1.718281828459046    3.078809711519162      1.791795536987222    54   1.718281828459046    3.078809711519162      1.791795536987222    55   1.718281828459046    3.078809711519162      1.791795536987222    56   1.718281828459046    3.078809711519162      1.791795536987222    57   1.718281828459046    3.078809711519162      1.791795536987222    58   1.718281828459046    3.078809711519162      1.791795536987222    59   1.718281828459046    3.078809711519162      1.791795536987222    60   1.718281828459046    3.078809711519162      1.791795536987222    61   1.718281828459046    3.078809711519162      1.791795536987222    62   1.718281828459046    3.078809711519162      1.791795536987222    63   1.718281828459046    3.078809711519162      1.791795536987222    64   1.718281828459046    3.078809711519162      1.791795536987222    65   1.718281828459046    3.078809711519162      1.791795536987222    66   1.718281828459046    3.078809711519162      1.791795536987222    67   1.718281828459046    3.078809711519162      1.791795536987222    68   1.718281828459046    3.078809711519162      1.791795536987222    69   1.718281828459046    3.078809711519162      1.791795536987222 jagy@phobeusjunior:~$  

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1

I'm sorry that this is not an answer, but it's worthwhile information that might help.

If we apply the Limit Comparison Test to the two series, putting the "harmonic" series below, we have the ratio $\left(\frac{n\,a_n}{a_1+\cdots+a_n}\right)^2$

Now if $a_n=f(n)$ where $f$ is an increasing continuous function, but not one that increases too quickly (as defined below when it matters) then

$\left(\frac{n\,a_n}{a_1+\cdots+a_n}\right)^2<\left(\frac{n\,f(n)}{\int_0^nf(x)\,dx}\right)^2$

And so if $f$ is slow-growing, as defined by $\int_0^nf(x)\,dx>C\,n\,f(n)$, then this ratio is bounded. So the Limit Comparison Test would give the convergence of $\sum\frac{n^2a_n}{(a_1+\cdots+a_n)^2}$.

I've found this problem to be much harder to tackle for quickly growing $a_n$, which is funny, since for these the series $\sum\frac{1}{a_n}$ has "more room" between it and a divergent series. If $a_n$ is all-the-time "quickly growing", then this lends itself to a direct examination of $\sum\frac{n^2a_n}{(a_1+\cdots+a_n)^2}$, where the denominator can be shown to be larger enough than the numerator to give convergence. But I think the real problem with any continued approach like this will be sequences that go back-and-forth between slowly growing and quickly growing.

And of course there is the concern that $a_n$ might not even be increasing.

  • 0
    Alex, I thought it was something like that. I went through the accepted answer line by line, more detail could be typed in at various points but it's correct.2012-09-25
1

Assume this fact $(\clubsuit)$: Prove that $\sum_{k=1}^n \frac{2k+1}{a_1+a_2+...+a_k}<4\sum_{k=1}^n\frac1{a_k}.$.

If you define $P_N=\sum_{n=1}^N a_n\,,\; C=\sum_{n=1}^{+\infty}\frac{1}{a_n}$ and $S_N=\sum_{n=1}^{N}\frac{n^2 a_n}{P_n^2}$ you have:

$S_N < \frac{1}{a_1} + \sum_{n=2}^{N}\frac{n^2(P_n-P_{n-1})}{P_n P_{n-1}} = \frac{5}{a_1}+\sum_{n=2}^{N-1}\frac{2n+1}{P_n}-\frac{N^2}{P_N},$

so, in virtue of $(\clubsuit)$, you have:

$S_N < \frac{2}{a_1} + 4C.$