# Questions tagged [ramanujan-summation]

Ramanujan summation is a technique invented by the mathematician Srinivasa Ramanujan for assigning a value to infinite divergent series.

44 questions
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### Help me find Taylor series for the function $\frac{1}{(-t;q)^2 _ \infty }$ [on hold]

$$\frac{1}{(-t;q)^2 _ \infty }$$ where $\frac{1}{(-t;q)_ \infty }$ q-pochhammer function
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### Convergence of series and summation methods for divergent series

I would like to know what is the sum of this series: $$\sum_{k=1}^\infty \frac{1}{1-(-1)^\frac{n}{k}}$$ with $$n=1, 2, 3, ...$$ In case the previous series is not convergent, I would like to know ...
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### If $(a,k)=(b,m)=1$, prove that $(ab,km)=(a,m)(b,k)$.

I'm reading the proof of the multiplicative property of $$s_k(n)=\sum_{d|(n,k)}f(d)g\bigg( \frac kd\bigg)$$ The book wrote that in order to understand the proof, we need to know if $a,b,k,m$ are ...
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### What consistent rules can we use to compute sums like 1 + 2 + 3 + …?

$\newcommand{ifelse}[3]{ \left( \begin{cases} #1\text{ if }#2 \\ #3\text{ otherwise} \end{cases} \right) }$ A recent Numberphile video on 1+2+3+... has made this question ("Why?") popular again, as ...
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### Is there any mathematical or physical situations that $1+2+3+\ldots\infty=-\frac{1}{12}$ shows itself? [duplicate]

I just saw the proof that $$1+2+3+\cdots=-\frac{1}{12}$$ and my brain still hurts. I completely understood the proof and my question is NOT actually about the proof itself. At the end of the proof, ...
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### Ramanujan 1918 paper

Does anybody know where I can find Ramanujan's paper from 1918 titled "On Certain Arithmetical Functions." It is referenced in wikipedia, under the Ramanujan Summation section, but I cannot find a ...
### Sum of the divisors of $n$, related to the Hardy-Littlewood circle method
Prove that $$\sigma(n) = \frac{\pi^2}{6}\,n\sum_{q = 1}^\infty q^{-2}c_q(n)$$ where $$c_q(n) = \sum_{a = 1, (a, q) = 1}^q \exp(2\pi i an/q)$$ and $\sigma(n)$ is the sum of the divisors of $n$. ...
Wikipedia states that Ramanujan sums and the Riemann Zeta function have the same values for even $k$: $$1 + 2^{2k} + 3^{2k} + \cdots = 0\ (\Re)$$ However, I don't understand how this can be true, ...