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Series sum approximation

Writing: NSum[(-1)^n/Sqrt[Log[n]], {n, 2, Infinity}, Method -> "AlternatingSigns"] I get: 0.690243 which is what you want. In particular, directly from ...
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A more convenient Fourier series

For the reasons mentioned above, I wrote the following "shell" for these functions: ...
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How to convert this term to a Hypergeometric function?

For your first question, if we gather the factors into a single variable z, there's a simple hypergeometric function: ...
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Help with Double Sum (lattice sum) over all integers m,n of 1/(a+m^2+n^2)

This kind of sum can be studied by integral transformation. Notice that $\int_0^1t^{x-1}dt=\frac{1}{x}$ if $\text{Re}(x)>0$: ...
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How to find the perturbation of $x^2 − 1 = \epsilon x$?

Decide up to which power you would like to expand: pow = 4; Let's do one of the equations you mentioned as an example (bring all terms to one side and save as a ...
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How to get the Taylor series of implicit functions

You can use AsymptoticSolve for this purpose: AsymptoticSolve[x+1/2y^2+1/2z+Sin[z]==0,{z,0},{{x,y},{0,0},4}] {{z -> -((2 x)/...
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Series vs Asymptotic in 12.1

Extended comment, I won't accept this as an answer. Here are some cases I've found where Series might be a better choice than ...
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How to represent $f(x) = (y-x)^k \log(y-x)$ as a summation of the form $f(x) = \sum\limits_{j=0}^\infty \cdots$?

Complete rewrite of answer The expression to be expanded as a series is the argument of ser = Series[(y - x)^k*Log[y - x], {x, 0, 5}] I attempted to obtain the ...
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Why can’t mathematica find this residue?

You could use SeriesCoefficient instead: SeriesCoefficient[(z+1)^2 Exp[3/z^2], {z, 0, -1}] 6 Addendum Another possibility ...
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Removing higher order terms

You can use a variation of the idea I gave here: Normal @ Series[ ss /. {f:u1|u2 -> (s f[#1,#2,#3,#4]&)}, {s, 0, 3} ] /. s->1 u2[x, y, z, t]^...
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What series does Mathematica use for Hypergeometric1F1?

Recall that $(-n)_k=0$ for $k>n, n,k\in\mathbb N$. Thus, what you have is an appropriate truncation of the usual series for the Kummer function. ...
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expand function as power series of another function

Try the following Series[f[InverseFunction[g][y]],{y,0,10}]
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