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I know a lot of techniques to show that a serie converge or diverge (Cauchy condensation test, Abel's test, comparision with an integrale, root test, ratio test...).

On the other hand I know nothing about caculating the value of a serie. Is there any theorems that may help ?

From now on I know the following :

  • integrale test can be combine with squeeze theorem
  • geometric serie

For example in this [How to prove $\lim \limits_{x \to 1^-} \sum\limits_{n=0}^\infty (-1)^nx^{n²} = \frac{1}{2} \ $? there are talking about :

  • Herglotz' trick
  • Weierstrass products.

  • ...

So I would like to know if there are some other nice theorem or slick techniques to find the value of a serie.

Thank you !

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In this regard, series are much like integrals: you need knowledge and, even more, ingenuity and experience.

Some basic Theorems, besides those you cited, are:
- Indefinite Sum;
- Summation by parts; - (Discrete) Convolution;
- Partial Fractions Decomposition;
- Binomial Tranform and others;
- Gosper's Algorithm and Hypergeometric Function;
- Abel's Binomial Theorem
and many others

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  • $\begingroup$ +1 for Gosper's Algorithm - found this a great presentation of why it works for WZ method: people.brandeis.edu/~gessel/homepage/slides/wilf80-slides.pdf @auhasard well word a read if you have a chance. Came across the method in a nice compendium called Irresistible Integrals think this is a free pdf: researchgate.net/publication/… plenty of material in here! $\endgroup$ – Mehness Jun 9 '18 at 16:39
  • $\begingroup$ (hoho pls ignore the bogus link, this is the book I meant if interested: Irresistible Integrals, Symbolics, Experiments, Evaluation $\endgroup$ – Mehness Jun 9 '18 at 21:17
  • $\begingroup$ @Mehness: thanks for indicating these interesting publications. I'll try and get through them. $\endgroup$ – G Cab Jun 9 '18 at 23:39
  • $\begingroup$ No problem sir, the book is definitely just sth to dip into if interested it's pretty hefty! $\endgroup$ – Mehness Jun 9 '18 at 23:48
  • $\begingroup$ @GCab Thank you ! This is actually what I was looking for, now it's time to train ! $\endgroup$ – auhasard Jun 10 '18 at 8:31
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I don't know if this is a good approach and if you mentioned this but it would add another possible approach which I posted on the linked question.

Consider

$$ F(y+1)-F(y) = e^{i\pi y} x^{y^2} = e^{i \pi y -ay^2} = f, $$

Take fourier transform of this and solve e.g. $$ \hat F(s) = \frac{\hat f(s)}{e^{is} - 1} $$ Note that $$ \lim_{n \to \infty} \sum_{i=0}^{n} F(i+1)-F(i) = - F(0) = \lim_{n \to \infty} F(n+1) - c \int \mathcal{F}(f)(s)\,ds = - c \int \mathcal{F}(f)(s)\,ds, $$ under the assumption $\lim_{n\to\infty} F(n+1) = 0$

For the example above it turns out the fourier transform turns into a delta measure as we approach $x\to 1$ and this might be an approach that is tractable and lead to convergence proof and limit calculation.

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  • $\begingroup$ I think you meant $n=\infty$ in your sum. Otherwise there's an extra term $F(n+1)$, . $\endgroup$ – Alex R. Jun 8 '18 at 23:32
  • $\begingroup$ thanks! I'll fix it $\endgroup$ – Stefan Jun 11 '18 at 15:09

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