8
What seems to help at least with NSum is increasing the number of terms that are computed explicitly via the option NSumTerms. This way one can get agreement for all first $100$ digits:
NSum[2^(-n + Sin[n Pi/5]), {n, 0, Infinity}, NSumTerms -> 400, WorkingPrecision -> 100]
2....
6
First determine for what value of $r$ the argument is less than $10^{-5}$:
η = 0.05211184484645051`;
sol = Solve[{η/(r + 1) + ArcTan[η/(r + 1)] == 1*^-5, r > 0}, r]
(* Out: {{r -> 10421.4}} *)
Then use that value as the upper limit in your summation:
Sum[η/(r + 1) + ArcTan[η/(r + 1)], {r, 0, Ceiling[r /. First@sol]}]
(* Out: 1.02436 *)
4
Sum[\[Eta]/(1 + r) + ArcTan[\[Eta]/(1 + r)], {r, 0, n}]
is given by
\[Eta]*EulerGamma - (1/2)*I*LogGamma[1 - I*\[Eta]] + (1/2)*I*LogGamma[1 + I*\[Eta]] +
(1/2)*I*LogGamma[2 - I*\[Eta] + n] - (1/2)*I*LogGamma[2 + I*\[Eta] + n] +
\[Eta]*PolyGamma[0, 2 + n]
The sum and the analytic expression diverge for n->Infinity.
For n > 10421 each additional ...
4
Clear["Global`*"]
To find the number of terms required, find the minimum value of n for which the last term is less than 10^-100
min = MinValue[{n, 2^(-n + Sin[n Pi/5]) < 10^-100}, n, Integers]
(* 334 *)
sum1 = NSum[2^(-n + Sin[n Pi/5]), {n, 0, Infinity}, NSumTerms -> 400,
WorkingPrecision -> 100]
(* 2....
4
Accumulate /@ matrix1
produces
{{a, a + b, a + b + c, a + b + c + d}, {e, e + f, e + f + g,
e + f + g + h}, {i, i + j, i + j + k, i + j + k + l}, {m, m + n,
m + n + o, m + n + o + p}}
3
Note, a double sum can be written with a single Sum. With this we make a replacement:
Sum[a[k], {k, 1, n}]*Sum[b[k], {k, 1, n}] /.
Sum[a1_, {a2_, 1, n}] Sum[b1_, {b2_, 1, n}] :>
Sum[(a1 /. a2 -> k1) (b1 /. b2 -> k2), {k1, 1, n}, {k2, 1, n}]
3
You can look at the Sin[] term in the exponent as some periodic noise. It would be good to smooth that out before trying numeric methods by summing over a period at a time:
sum = NSum[
Sum[2^(-(10 n + k) + Sin[(10 n + k) Pi/5]), {k, 0, 9}],
{n, 0, Infinity}, WorkingPrecision -> 100]
(*
2....
1
Instead of evaluating a single sum, you could break up the sum into two or more sums.
L = 24;
sind = Range[-Pi, Pi, 2*Pi/L];
a[x_, y_] := x y
g[x_, y_] := x + y
f = Cos[x]*Sin[y]*Sin[x + x1]*Cos[y + y1]*a[x, y] + g[x, y]*Cos[x2 - y2]*Sin[x + x1 + x2];
AbsoluteTiming[sx = Sum[f, {x, sind}, {x1, sind}, {x2, sind}] // Simplify;
sxy = Sum[sx, {y, sind}, {y1, ...
1
Posting as an answer at the request of the OP
There is probably no built-in way to achieve what you want because the result you are after is in general not correct: in order to get it you need to rewrite and reorder the summations, which is only possible under special convergence conditions.
That being said, if you are sure beforehand that all your ...
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