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I'm working a lot with series these days, and I would like to know if there are any texts, papers, articles that might suggest a general outline for finding $n$th partial sums of convergent series. Most of my searching turns up methods for finding the sums of geometric/telescoping/power series etc., but I'd like to know if there are any general guidelines that are followed for finding partial sums for something like $$\sum_{k=1}^{\infty}\frac{1}{k^2}$$ or $$\sum_{k=1}^{\infty}\frac{6^k}{(3^{k+1}-2^{k+1})(3^k-2^k)}$$.

I've seen solutions to both of these, and they're beautiful and unintuitive. So I wonder what, if any, methods might be used to get an edge on finding their $n$th sums.

Aside from listing partial sums and looking for patterns, what approaches do mathematicians commonly use to solve problems like this? Or, if listing partial sums is the best route to take, what can or should be done to improve pattern recognition?

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An excellent resource for learning how do do such things is Concrete Mathematics, by Graham, Knuth and Patashnik. Section 5.8, on Mechanical SUmmation and Gosper's method, gives a very general way to attack such problems.

However, be aware that many such partial sums have no nice closed form, even when the infinite sum does have a nice closed form. For example, while $$\sum_1^\infty \frac1{n^2} = \frac{\pi^2}6$$ I know of no nice form for $$H_{2,k} = \sum_1^k\frac1{n^2} $$ (other than the fact that people have named this function the harmonic number of order 2).

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You might like to explore (if you haven't already) the topic of Generating Functions as there is this neat result:

Given a sequence {a} with a known generating function, $GF$, then the sequence of partial sums of sequence {a} is given by; $$GF \times \frac{1}{1-x}$$ So, for example, the Fibonacci Sequence has the the generating function, $$\frac{x}{1-x-x^2}=0+1x+1x^2+2x^3+3x^4+5x^5+8x^6+...$$ And the partial sums of the Fibonacci Sequence has the generating function, $$\frac{x}{(1-x)(1-x-x^2)}=0+1x+2x^2+4x^3+7x^4+12x^5+20x^6+...$$

A lot of the time, as you are only dealing with a formal power series, you do not need to worry about convergence issues. Also, you can continue to work with the sequence even when there is no closed form formula for the $n^{th}$ term, via the generating function.

In response to Mark Fischler's comment below the closed form formula for the partial sums of the Fibonacci Sequence are not that difficult to get via generating functions. To begin, use partial fractions to obtain, $$\frac{x}{(1-x)(1-x-x^2)}=\frac{1+x}{1-x-x^2}-\frac{1}{1-x}$$ The $-\frac{1}{1-x}$ is the $-1$ on the end of Mark's formula so we can now focus on forcing the other fraction to factorise further, $$\frac{1+x}{(1-x-x^2)}=\frac{1+x}{\big(1-\frac{1-\sqrt{5}}{2}x\big)\big(1-\frac{1+\sqrt{5}}{2}x\big)}$$ Another dose of partial fractions gets us to, $$\frac{1+x}{(1-x-x^2)}=\frac{5+3\sqrt{5}}{10} \times \frac{1}{\big(1-\frac{1+\sqrt{5}}{2}x\big)}+\frac{5-3\sqrt{5}}{10} \times \frac{1}{\big(1-\frac{1-\sqrt{5}}{2}x\big)}$$ These are all standard bits which translate directly into the formula that Mark gives, not forgetting to add that $-1$ back in on the end.

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  • $\begingroup$ Fine, but nowhere does the GF for the partial sum sequence $S_n = \{0,1,2,4,7,12,20,33,\ldots\}$ make it easy to see the closed form $$S_n = \left(\frac12+\frac3{10}\sqrt{5}\right)\left(\frac12(1+\sqrt{5})\right)^n + \left(\frac12-\frac3{10}\sqrt{5}\right)\left(\frac12(1-\sqrt{5})\right)^n -1$$ $\endgroup$ – Mark Fischler Mar 20 at 18:01
  • $\begingroup$ @Mark Fischler : It's not too bad, and all standard techniques provided care is taken with the algebra. Rather than just 'talk the talk' I've added how to get the closed form formula you give with generating functions to the bottom of my posted answer (as it was too long to go in a comment). I enjoyed writing it out and will use it as an exercise question next time I teach this topic. $\endgroup$ – Martin Hansen Mar 20 at 21:42
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    $\begingroup$ Nice! +1 for presenting a useful approach I had not thought of! $\endgroup$ – Mark Fischler Mar 21 at 2:31

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