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I'm exploring the possibilities of the Limit function:

Example 1:

$\lim_{x\to \infty } \, \frac{\sqrt{x-1}}{x}$

Limit[Sqrt[x - 1]/x, x -> Infinity]

0

Example 2:

$\lim_{n\to \infty } \, \frac{(n!)^{1/n}}{n}$

Limit[n!^(1/n)/n, n -> Infinity]

$1/E$

But when I tried to use a summation limit it did not work out:

$\lim_{n\to \infty } \, \left(\sum _{a=1}^n a^{n-a}\right)$

Limit[Sum[a^(n - a), {a, 1, n}], n -> Infinity]

The above idea was to describe this summation:

$1^{10-1}+2^{10-2}+3^{10-3}+4^{10-4}+5^{10-5}+6^{10-6}+7^{10-7}+8^{10-8}+9^{10-9}+10^{10-10}+\text{...}+\infty ^{10-\infty }$

Is it possible to get some numerical result? Maybe using NSolve, Solve ....

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  • $\begingroup$ Limit is for Real-valued variables, not Integer-valued (as you use). $\endgroup$ – David G. Stork Jul 13 '17 at 17:25
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First, we note all the terms of the sum under cosideration are positive. Second, we take one of them, namely

Floor[n/2]^(n - Floor[n/2])

, and consider its limit as $n$ approaches $\infty$:

Limit[Floor[n/2]^(n - Floor[n/2]), n -> Infinity]

$\infty$.

This implies the limit under consideration is infinite too. Of course, there may exist a generalised sum of the series under consideration, but this is a math stuff, not a Mathematica matter.

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  • $\begingroup$ +1, good approach. But why not just take the first term in the sum, since that is the largest and dominant term? You get same result of course. The sum blows up. $\endgroup$ – Nasser Jul 13 '17 at 18:22
  • $\begingroup$ @Nasser: The first term is $1^{n-1}=1$ and the last term is $n^0=1$. $\endgroup$ – user64494 Jul 13 '17 at 18:27
  • $\begingroup$ I see. good point. $\endgroup$ – Nasser Jul 13 '17 at 18:29

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