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Is there a way to find precise asymptotics or better bounds of series such as $\sum_{n=1}^{\infty}x^{n+1/n}/n!>e^x$ ?

Or $\sum_{n=1}^{\infty}x^{\sqrt n}/e^n$?

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Where did that question come from? – Patrick Da Silva Apr 24 '12 at 19:55
Straight from my mind – TROLLKILLER Apr 24 '12 at 20:16
You know, asking yourself a random question can be very painful to your brain sometimes ; I'm not saying this is a bad question, perhaps there is an answer that would be nice for this question, but what I'm saying is ; there exists a ton of questions in this world, way more than there exists answers. When choosing the question you wish to spend time on you should be aware of the fact that there are maybe more interesting questions you could ask yourself that are out there waiting for your brain to bash on them. If there is no reason why you're asking yourself those questions – Patrick Da Silva Apr 24 '12 at 20:56
, then I suggest either that you find a reason or that you think about something else to worry about (in the general case). Now if someone finds an answer to this, good for you, I'll be happy =) But next time think about this ; it'll make you a better mathematician. – Patrick Da Silva Apr 24 '12 at 20:57
@Administrator Is it $n+{1 \over n}$ or ${n+1 \over n}$ in the exponent? – Pedro Tamaroff Apr 24 '12 at 21:41
up vote 2 down vote accepted

In the first case, the sums $S(x)$ are such that $\exp(-x)S(x)\to1$ when $x\to+\infty$. In the second case, the sums $T(x)$ are such that $\exp(-(\log x)^2/4)T(x)\to2\sqrt{\pi}$ when $x\to+\infty$.

A nice probabilistic interpretation helps in the first case while some raw real analysis techniques inspired by Laplace's method solve the second case. The two questions do not have much in common. No idea what "better bounds" means.

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And the absence of comment by the OP is puzzling. – Did Jul 29 '14 at 13:47

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