# What is the limit of the sum as t approaches infinity?

I am unable to find the limit of the following sum as t appraoches infinity

Clear["Global*"]
S[t_] := S[
t] = (1/(t - 1)) Sum[
Sum[Abs[(t)/((t - 1) n (n + 1)) - 1/(t - 1)], {n, 1, j}], {j, 1,
t - 1}]


As t grows larger the computation time of the sum slows down.

How do we compute the limit?

Edit: According to the comments I tried RealAbs and computed the limit as t->Infinity

Clear["Global*"]
S[t_] := S[
t] = (1/(t - 1)) Sum[
Sum[RealAbs[(t)/((t - 1) n (n + 1)) - 1/(t - 1)], {n, 1, j}], {j,
1, t - 1}]
N[Limit[S[t], {t -> Infinity, Assumptions -> Element[t, Reals]}]]



However, the computation time takes forever to compute.

• If t is real, try adding it as an assumption and using RealAbs instead of Abs. Commented Sep 28, 2021 at 15:50
• @MichaelE2 I tried using RealAbs and Limits but my sums still takes a long time to compute. Commented Sep 28, 2021 at 16:08

We can make an asymptotic approximation by hand and take its limit:

sum = (1/(t - 1)) Sum[    (* takes ~80 sec *)
RealAbs[(t)/((t - 1) n (n + 1)) - 1/(t - 1)], {j, 1, t - 1}, {n,
1, j}, Assumptions -> t \[Element] Reals];

asymp = FullSimplify[sum /. Ceiling | Floor -> Identity, t > 100] /.
f : _PolyGamma | _HarmonicNumber :>
With[{res = Normal@Series[f, {t, Infinity, 3}]}, res /; True] //
Simplify;

Limit[asymp, t -> Infinity]
(* 3/2 *)


Numerical evidence for the validity of the approximation. I'll let someone else prove it rigorously.

TableForm[
Table[Style[sum - asymp, PrintPrecision -> 16] // AbsoluteTiming, {t,
10.32^Range@10}],
]


Continuation of the fine answer of @MichaelE2

sum = (1/(t - 1)) Sum[
Abs[(t)/((t - 1) n (n + 1)) - 1/(t - 1)], {j, 1, t - 1}, {n, 1,
j}, Assumptions -> t \[Element] Reals] // Simplify[#, t > 2] &

(*   (-4 (-1 + t) Floor[1/2 (-1 + Sqrt[1 + 4 t])]^2 +
2 Floor[1/2 (-1 + Sqrt[1 + 4 t])]^3 +
t (1 - 2 EulerGamma + t -
2 PolyGamma[0, 2 + Floor[1/2 (-1 + Sqrt[1 + 4 t])]]) +
Floor[1/2 (-1 + Sqrt[1 + 4 t])] (2 - 3 t - 2 EulerGamma t +
3 t^2 - 2 t PolyGamma[0,
2 + Floor[1/2 (-1 + Sqrt[1 + 4 t])]]))/(2 (-1 + t)^2 (1 +
Floor[1/2 (-1 + Sqrt[1 + 4 t])]))   *)


The argument of all Floor is the same and it goes to infinity as t goes to infinity. It is sufficient to regard only cases where this is an integer and therefore Floor->Identity.

solt = First@Solve[1/2 (-1 + Sqrt[1 + 4 t]) == k, t, Integers] //
Simplify[#, {t \[Element] Integers, k \[Element] Integers, k > 2}] &

(*   {t -> k (1 + k)}   *)

sum2 = sum /. solt // Simplify[#, k \[Element] Integers && k > 2] &

(*   (k (1 + k) (3 - 2 EulerGamma - 3 k + 3 k^2 -
2 PolyGamma[0, 2 + k]))/(2 (-1 + k + k^2)^2)   *)

Limit[sum2, k -> \[Infinity]]

(*   3/2   *)


Do both divisions by (t-1) belong in the expression? If not, it must diverge. To see that, lay out the terms for any t as a lower triangular matrix. E.g.,

With[{t = 5}, PadRight@
Table[Abs[t/(n (n + 1)) - 1]/(t - 1), {j, 1, t - 1}, {n, 1, j}]] // MatrixForm


$$\left( \begin{array}{cccc} \frac{3}{8} & 0 & 0 & 0 \\ \frac{3}{8} & \frac{1}{24} & 0 & 0 \\ \frac{3}{8} & \frac{1}{24} & \frac{7}{48} & 0 \\ \frac{3}{8} & \frac{1}{24} & \frac{7}{48} & \frac{3}{16} \\ \end{array} \right)$$

All the entries must be positive. In any column, all the entries are the same. Now look at the first column. It's entries are all (t/2 - 1)/(t-1)`, which has a limit of (1/2). If the double-deflation by $$(t-1)$$ was intentional, then after weighting this column has a limiting sum of $$1/2$$, which puts a floor under the total.

• Numerically, I'm getting a limit of 3/2 (up to $t=10^{12}$). Commented Sep 28, 2021 at 16:29
• @MichaelE2 Ah, I failed to deflate by (t-1).
– Alan
Commented Sep 28, 2021 at 16:48