I want to check the following, Hardy's most fundamental inequality, by using Mathematica:

$$\sum_{n=1}^\infty \left(\frac{A_n}{n}\right)^p<\left(\frac{p}{p-1}\right)^p\sum_{n=1}^\infty a_n^p$$ where $p>1,$ $a_n\geq0$ and $A_n=a_1+a_2+\cdots+a_n$

I tried:

hardy[l_List, p_] := Module[{i, n = Length[l]},
Sum[(Sum[l[[i]], {i, n}]/n)^p, {n, 1, Infinity}] < 
(p/(p - 1))^p Sum[l[[i]]^p, {n, 1, Infinity}]]

hardy[Table[1/n^2, {n, 1, Infinity}], 2]

but there is an error message.

I want to do something like this:

define random sequence randomSeq $a_n$, then


sum1[p_,randomSeq_]:=$\sum_{n=1}^\infty \left(\frac{A_n}{n}\right)^p$

sum2[p_,randomSeq_]:=$\left(\frac{p}{p-1}\right)^p\sum_{n=1}^\infty a_n^p$


Plot[{sum1[p,randomSeq], sum2[p,randomSeq]}, {p, 2, 100}]
  • $\begingroup$ When you do Table[1/n^2, {n, 1, Infinity}] you are asking MMA to make an infinitely long array. It can't do that of course. You can try to take the difference between the LHS and RHS of the inequality and plot it for larger and larger a[n]-tables, and see whether it tends to something negative. $\endgroup$ Commented Jun 9, 2016 at 21:19

1 Answer 1


Here's a solution if you're willing to truncate the infinite sums.

(* a deterministic random sequence... *)
a[n_] := a[n] = BlockRandom[SeedRandom[n]; RandomReal[]/n]

A[n_] := A[n] = Total[a /@ Range[n]]

lhs[p_?NumericQ, cap_] := Sum[(A[n]/n)^p, {n, 1, cap}]

rhs[p_?NumericQ, cap_] := (p/(p - 1))^p Sum[a[n]^p, {n, 1, cap}]

Plot[{lhs[p, 1000], rhs[p, 1000]}, {p, 1, 4}]

enter image description here


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