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1

Thanks to @J. M. this solution gives the answer: Sum[a^i, {i, ∞}, GenerateConditions -> True] It returns a ConditionalExpression, in the above case: ConditionalExpression[-(a/(-1 + a)), Abs[a] < 1]


3

As an example with m = 5, n = 3, and f[ki][x] the functions in the question, the Sum can be written as m = 5; Sum[KroneckerDelta[m - (k1 + k2 + k3)] Multinomial[k1, k2, k3] f[k1][x] f[k2][x] f[k3][x], {k1, m}, {k2, m}, {k3, m}] (* 90 f[1][x] f[2][x]^2 + 60 f[1][x]^2 f[3][x] *) See the documentation for the definition of Multinomial.


7

Reverse the order of the summation. i.e., k -> (n - k + 1) s = Sum[(n - k + 1)/(n^2 - (n - k + 1) + 1), {k, 1, n}] // Simplify (* -n + (1 + n^2) PolyGamma[0, 1 + n^2] - (1 + n^2) PolyGamma[0, 1 - n + n^2] *) Limit[s, n -> Infinity] (* 1/2 *) For an alternative representation s2 = FullSimplify[s] (* -n - (1 + n^2) HarmonicNumber[(-1 + n) n] ...


3

In cases where you can't get a symbolic result, it's also possible to use a completely numerical approach: Needs["NumericalCalculus`"] sum[n_?NumberQ] := NSum[k/(n^2 - k + 1), {k, 1, n}] NLimit[sum[n], n -> Infinity] (* ==> 0.499999 *)


8

You can use the Euler-Maclaurin formula to get the limit (the sum can be approximated by an integral, which becomes exact in the infinite limit): f[i_] = i/(n^2 - i + 1); Integrate[f[k], {k, 0, n}, Assumptions -> n > 0] Limit[%, n -> Infinity] 1/2


1

The problem is that without the limit, the sum doesn't converge, and without the sum the limit is 0. Mathematica can only do one at a time. At a finite value n, the sum gives a sequence with a $\frac{1}{0}$ or other complexinfinity expression for all values of n, which is a by product of the sum not converging. You can approximate the limit: i=4; Sum[k/(n^...


3

It seems to help to explicitly expand Abs, which by default is a function over the complexes, for a real/integer argument: Assuming[ {k, m} ∈ Integers, Sum[t^Abs[k - m] // PiecewiseExpand, {k, -∞, ∞}, Assumptions -> m ∈ Integers && Abs[t] < 1] ] (* (-1 - t)/(-1 + t) *) Follow-up to comment: Maybe deconstruct the pieces semi-...


5

This sum is indeed possible to compute as it is using the FourierSequenceTransform. Some algorithms of MA are obviously more advanced than others: s = FourierSequenceTransform[t^Abs[n - m], n, w, Assumptions -> m \[Element] Integers] (*(E^(-I (-1 + m) w) (-1 + t^2))/((E^(I w) - t) (-1 + E^(I w) t))*) Now we recall that for $w\rightarrow 0$ the desired ...


3

Try changing the method of summation: Clear["Global`*"] β = 1; d = 12; V = 1/4; Σf = 1 + I; ω[m_] := ((2*m + 1)*π) ν[n_] := (2*n*π)/β F1[n_] := NSum[(V^2*8)/ d^2*(I*ω[m] - I*Sign[ω[m]]* Sqrt[(ω[m])^2 + (d/2)^2])*1/(I*ω[m] + I*ν[n] - Σf), {m, -Infinity, Infinity}, Method -> "WynnEpsilon", WorkingPrecision -> 30] 0....


2

f1[n_] = Sum[Fibonacci[i]*(-1)^i, {i, 0, n}]; If you know or suspect the general form of the model model = (-1)^n Fibonacci[b*n + c] + d; data = Table[f1[n], {n, 10}] // Simplify; f2[n_] = model /. ( NonlinearModelFit[data, model, {b, c, d}, n][ "BestFitParameters"] /. x_?NumericQ /; (Abs[x - Round[x]] < 10^-10) :> Round[x]) (* -1 + ...


4

There is probably no package or algorithm that can give you answer in all cases. However, I would like to suggest a semi-automatic solution. I will illustrate it with few examples. Notice, I do not know solutions in advance. Example 1 t[1] = Table[Sum[Fibonacci[3 i], {i, 0, k}], {k, 0, 10}] *({0, 2, 10, 44, 188, 798, 3382, 14328, 60696, 257114, 1089154}*) ...



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