I was dwindling with the Basel series and thinking of ways to prove it. I did this: \[ \displaystyle \sum_{n=0}^{\infty} \dfrac{1}{n^{2}+1} = \int_{0}^{1} \sum_{n=0} ^{\infty} x^{n^{2}}dx \] So, the series is: \[ \displaystyle \sum_{n=0}^{\infty} x^{n^{2}}\] for \( |x| <1 \) Wolfram Alpha gives the result in terms of Jacobi Theta Functions, which I currently fail to comprehend, So, what are your views on it??

And, more generally for the series: \[ \displaystyle \sum_{n=0}^{\infty} x^{n^{\alpha}}\] where, \(\alpha\) is a positive integer or maybe a real, and \(|x|<1\)

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TopNewestI don't know why you want to convert it into an integral (which is wrong by the way). You can simply solve it via digamma functions or residues to get the answer of \( \frac12(\pi \coth(\pi) + 1) \). – Pi Han Goh · 1 year, 10 months ago

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@Pi Han Goh It's not about the sum provided in the beginning, it's about the series given at the end – Upanshu Gupta · 1 year, 10 months ago

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@Kartik Sharma @Michael Mendrin @Pi Han Goh @Ishan Singh – Upanshu Gupta · 1 year, 10 months ago

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@Ishan Singh Well I was just calculating \(\zeta(2)\) using the aforesaid method, when I encountered the series. – Upanshu Gupta · 1 year, 10 months ago

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BTW, you can use Laplace transform too. – Kartik Sharma · 1 year, 10 months ago

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@Kartik Sharma I got: \[ \displaystyle \int_{0}^{\infty} \dfrac{\sin(mx)dx}{1-e^{-x}} = \dfrac{1}{2m} + \dfrac{\pi}{2}\coth(\pi m) \] – Upanshu Gupta · 1 year, 10 months ago

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– Kartik Sharma · 1 year, 10 months ago

Yeah. BTW, yeah, I am thinking about the series only for a moment. I too have not yet got any chance to enjoy Theta functions but have heard a lot about them. I would wish to learn about them at the end of my course with Wilfred Kaplan's book(here course means self study of Kaplan). But yeah, I wish to think of these without any theta.Log in to reply

@Kartik Sharma Sorry, for mingling the prob like this , but what I want in particular is the summation of that particular series and that if it can be put up in a closed form other than that of jacobi functions.My point is not emphasized on Jacobi functions, but the "series" – Upanshu Gupta · 1 year, 10 months ago

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– Pi Han Goh · 1 year, 10 months ago

I don't mean to ruin your fun but if there's a simpler way to evaluate \( \displaystyle \sum_n q^{n^2} \) without using Jacobi Theta functions, mathematicians would have known about it already.Log in to reply

Ramanujan "guessed" it, why can't we? (This is based on the fact that many people believe that Ramanujan guessed almost every one of his results ans didn't provide results... He should not be regarded as a true mathematician(sarcasm of course)). – Kartik Sharma · 1 year, 10 months ago

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The only thing that I can think of is Rogers-Ramanujan Identities but I doubt that's even relevant to the question at hand.

Yeah, Pierre de Fermat is not a mathematician either (sarcasm). – Pi Han Goh · 1 year, 10 months ago

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– Kartik Sharma · 1 year, 10 months ago

Okay, now the following is random but I just wanted to know more about Skewes' number and more about the shift of inequality sign of \(\pi(x)\) and \(\text{Li}(x)\), can you help?Log in to reply

video already. I only know as much as you.

I assumed you've watched thisWhether Riemann hypothesis is true or not, the inequality \(\pi(n) < \text{li}(n) \) will fail for a number lower than \(e^{e^{e^{79}}} \) or for a number larger than \(10^{10^{10^{10^3}}} \). Since I'm not verse in Logarithmic Integral, I doubt I can push further than this. – Pi Han Goh · 1 year, 10 months ago

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