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1

Truncate early and often. f[0] = f0; f[n_] /; n > 0 := f[n] = Sum[ Series[1/(6 + j x) n (1/x + j f[n - 1]), {x, 0, 0}], {j, 1, 4}]; Timing[ res = Series[ Sum[Series[1/(2 + n x) (n/x + f[n]), {x, 0, 0}], {n, 1, 6}], {x, 0, 0}]] (* Out[140]= {0.012000, SeriesData[x, 0, { Rational[1016635, 162], Rational[5, 486] (-3351289 + 835701 f0)}, ...


2

Depending on your specific needs the straightforward way to do this is Series[p[x+e,y-e,t],{e,0,1}] or Series[p[x+ex,y-ey,t],{ex,0,1},{ey,0,1}] If you want to extract the terms proportional to $\epsilon$ you can get the correct coefficients with SeriesCoefficient[p[x + e, y - e, t], {e, 0, 1}] or if that suits you better with ...


4

Partial sums of this sequence are given by: Sum[(-1)^(n + 1) Cos[3^n x]^3/3^n, {n, 1, m}] (* output: 1/4 Cos[3 x] - 1/4 (-(1/3))^m Cos[3^(1 + m) x] *) For real $x$, we know this converges because $\cos(3^{1+m}x)$ is bounded. Mathematica does not assume $x$ is real and, as Bob Hanlon notes, will produce the correct result by evaluating a partial sum, ...


1

There is a problem with the entry for terms. Retype the entry and it will work properly as below. terms = 1 + Accumulate[2. Table[1/(1 + (2 n)^2), {n, 1, 500}]]; n = {50, 100, 200, 500}; p = terms[[n]]; exact = \[Pi]/2. Coth[\[Pi]/2.]; pd = 100. (exact - p)/exact; TableForm[{{n, p, pd}}, TableHeadings -> {None, {"N", "s", "% diff"}}]


6

In Mathematica 10 you can use SeriesCoefficient and Inactive to get what you require Inactive[Sum][SeriesCoefficient[2/(3(x-1)^3),{x,0,n},Assumptions->n>=0]x^n,{n,0,\[Infinity]}] where Inactive prevents Sum from evaluating. You can then "activate" the Sum as follows Activate[%] to get back to your original expression.


2

This is your expression: expr1 = 1/(x + 1); This is the the change of variables: sl = Solve[x - 1 == z, x][[1, 1]] (* x -> 1 + z *) Substituted to the expression it yields this: expr2 = expr1 /. sl (* 1/(2 + z) *) Now it can be expaneded as you need: Series[expr1 /. sl, {z, Infinity, 5}] /. z -> HoldForm[x - 1] (* SeriesData[ ...


3

You can expand around infinity as follows: Series[x/(1 - 1/x), {x, Infinity, 2}] $x + 1 + \frac{1}{x} + \left(\frac{1}{x}\right)^2 + \mathcal{O}\left[\left(\frac{1}{x}\right)^3\right]$ This returns a SeriesData expression, so you might want to add in a call to Normal as b.gatessucks suggests.



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