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I am trying to evaluate the following limit of an oscillating sum,

$$ S=\lim_{x \to \infty}\sum _{i=3}^{\infty} \frac{(-1)^{i-1} x ^{i-\frac{4 (i-1)}{i+1}}}{(i-1)!} $$ which looks like

$$ \lim_{x \to \infty} \left(\frac{x }{2} - \frac{x ^{8/5}}{6} + \frac{x ^{7/3}}{24} - \frac{x ^{22/7}}{120} + \dots \right) $$

I'm trying to use summation by parts, which is the method one is often meant to use for oscillating sums, but does this lead anywhere? Is there a way to get the limit of the oscillating sum otherwise?

What is the right approach for limits of oscillating sums like this? Perhaps bringing the limit inside the sum?

apg
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  • Using summation by parts leads I think to $$S=\lim_{x \to \infty} \left[\lim_{N \to \infty}\left(\left(\sum_{i=0}^{N}(-1)^i\right)\frac{x ^{N-\frac{4 (N-1)}{N+1}}}{(N-1)!}-\sum_{i=3}^{N-1}\left[\left(\sum_{j=0}^{N}(-1)^j\right)\left(\frac{x ^{i-\frac{4 i}{i+2}+1}}{i!}-\frac{x ^{i-\frac{4 (i-1)}{i+1}}}{(i-1)!}\right)\right]\right)\right].$$ – apg Aug 26 '22 at 22:18
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    The sum is equal to $$F(x) = \frac 1 {2 \pi i x^3} \int_{(-\infty - i, -3/2 - i, -3/2 + i, -\infty + i)} \Gamma(s) e^{\phi(s) \ln x} ds$$ with $\phi(s) = -s - 8/(s - 2)$. By moving the contour to the right, we can show that $F(x)$ is equal to minus the sum of the residues at $s = -1, 0, 2$ plus $o(x^p)$ for any $p$. To estimate the sum of the residues at $s = 0, 2$, we apply the steepest descent method: $$\int_{|s - 2| = 2 \sqrt 2} \Gamma(s) e^{\phi(s) \ln x} ds \sim -i \int_{\mathbb R} \Gamma(s_0) , e^{(\phi(s_0) - \phi''(s_0) \xi^2/2) \ln x} d\xi$$ with $s_0 = 2 - 2 \sqrt 2$. – Maxim Sep 02 '22 at 18:32
  • Can I ask, what method is this? Is it common to use complex analysis for oscillating sums? – apg Sep 02 '22 at 22:07
  • @Maxim, similar to this? https://math.stackexchange.com/questions/1746711/alternating-series-test-for-complex-series – apg Sep 02 '22 at 22:13

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