Contest problem: Show $\sum_{n = 1}^\infty \frac{n^2a_n}{(a_1+\cdots+a_n)^2}<\infty$ s.t., $a_i>0$, $\sum_{n = 1}^\infty \frac{1}{a_n}<\infty$
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.
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 $$ \sum_{k=1}^\infty\frac{p_k}{\bar{p}_k^2}\le20\sum_{j=1}^\infty\frac1{p_j}\tag{8} $$
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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