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Well it's me, Marko Riedel. ;-)

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2d
revised Finding all possible combination **patterns** - as opposed to all possible combinations
fix typo
2d
answered Finding all possible combination **patterns** - as opposed to all possible combinations
Jan
22
answered Curious Binomial Coefficient Identity
Jan
21
comment Closed form as sum and combinatorial of Fibonacci numbers
It seems you did not distinguish between your summation index and the coefficient extractor. It is not necessary to calculate anything else here, just use CIF. Good luck, MSE says to conclude.
Jan
21
comment Closed form as sum and combinatorial of Fibonacci numbers
I don't quite follow. Maybe you should typeset and post your work (without altering the intent of the question).
Jan
21
comment Closed form as sum and combinatorial of Fibonacci numbers
There is an example of the technique at this MSE link I and another example at this MSE link II. The title refers to a classic text on generating functions by H. Wilf which I believe is available for download on his website.
Jan
21
comment Closed form as sum and combinatorial of Fibonacci numbers
I believe Wilf / generatingfunctionology is the easiest here, it does not use complex variables. The first component is zero because there is no pole inside the circle $|z|=\epsilon.$
Jan
21
answered Closed form as sum and combinatorial of Fibonacci numbers
Jan
21
comment How closely can we estimate $\sum_{i=0}^n \sqrt{i}$
This sum is investigated at this MSE link.
Jan
20
answered Hypergeometric 2F1 with negative c
Jan
20
answered Labelling the Vertices of Dodecahedron
Jan
17
answered Showing ${n + a - 1 \choose a - 1} = \sum_{k = 0}^{\left\lfloor n/2 \right\rfloor} {a \choose n-2k}{k+a-1 \choose a-1}$
Jan
17
revised Trees with odd degree sequence
treat the labeled case
Jan
17
revised Trees with odd degree sequence
multisets rather than sets II
Jan
16
answered Trees with odd degree sequence
Jan
13
comment Finding the coefficients of $h(z)$ laurent series
I apologize for not having seen this sooner. I would say learn about the so-called Cauchy product which is a very simple concept yet part of the syllabus here in Germany. If you write down the series for $\psi(-z)/(z+2)^3$ and $1/(1+z)$ then the Cauchy product will let you read off the coefficient on $(z+2)^{-1}.$ This is also at Wikipedia.
Jan
12
revised $\sum_{n=1}^{\infty} \frac{1}{(n+1)(n+2)^3}$ using complex analysis
remember
Jan
12
revised $\sum_{n=1}^{\infty} \frac{1}{(n+1)(n+2)^3}$ using complex analysis
exponent
Jan
12
answered Closed form solution and combinatorial proof.
Jan
12
answered $\sum_{n=1}^{\infty} \frac{1}{(n+1)(n+2)^3}$ using complex analysis