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 !


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

  • $\begingroup$ +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

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.


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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.