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I read about this https://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula which is Euler-Maclaurain and I did my calculation to find the constants. However, the answer differ greatly from expected value. Are there any other series which is more compatible to deal with sum of n^n so that we get a nice closed form?

Sum of 1.5-powers of natural numbers

Here is an example but using sum of $n^{1.5}$ but how about approximating a closed form for sum of $n^n?$

If there is a nice closed form (a short one), it would be awesome.

Thomas Andrews
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Maximus Su
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    Wouldn't that require solving $\int_0^n x^x , dx$? As far as I know that integral can only be expressed in terms of another series, which kind of defeats the point. – Michael Seifert Jun 29 '21 at 17:45
  • @MichaelSeifert You can expand the x^x using taylor series and integrate maybe 3 first terms individually. That's what i did because first 3 terms should have good accuracy already. A – Maximus Su Jun 29 '21 at 17:47
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    Would something like $$ n^n \left( {1 + \frac{1}{{ne}} + \mathcal{O}!\left( {\frac{1}{{n^2 }}} \right)} \right) $$ be sufficient? – Gary Jun 29 '21 at 17:49
  • @Michael Seifert Hi professor, if there is a closed form for this and yet accurate, that would be so awesome – Maximus Su Jun 29 '21 at 17:49
  • @Gary The values are close but can that accuracy be increased further? Any links i can refer to? – Maximus Su Jun 29 '21 at 17:51
  • I don't see how the sort of Taylor series approach you're describing would work. If you're expressing$$x^x = \sum_{m=0}^\infty \frac{(x \ln x)^m}{m!}$$and truncating the series at the $M$th term for some fixed $M$, then the remainder term will be of the order $(x \ln x)^M/M!$. If you then integrate this from 0 to $n$, the integral of this remainder term will get larger as $n$ increases, and your approximation would lose accuracy. – Michael Seifert Jun 29 '21 at 17:58
  • @Michael Seifert Hi sir, my mistake sorry :) – Maximus Su Jun 29 '21 at 17:59
  • The first three terms are not accurate, no. Consider $e^x.$ The first three terms are not a good representation og $e^x$ for $x$ large. – Thomas Andrews Jun 29 '21 at 18:00
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    How about $$ \sum\limits_{k = 1}^n {k^k } \sim n^n \left( {1 + \frac{1}{{en}} + \left( {\frac{1}{{2e}} + \frac{1}{{e^2 }}} \right)\frac{1}{{n^2 }} + \left( {\frac{7}{{24e}} + \frac{2}{{e^2 }} + \frac{1}{{e^3 }}} \right)\frac{1}{{n^3 }} + \cdots } \right)? $$ – Gary Jun 29 '21 at 18:13
  • @Gary Wow, that s so accurate! May i know what concept is this and how you derived this? – Maximus Su Jun 29 '21 at 18:14
  • Starting from the "end" of the series, $$ \sum\limits_{k = 1}^n {k^k } = n^n \left( {1 + \frac{1}{n}\left( {\frac{{n - 1}}{n}} \right)^{n - 1} + \frac{1}{{n^2 }}\left( {\frac{{n - 2}}{n}} \right)^{n - 2} + \frac{1}{{n^3 }}\left( {\frac{{n - 3}}{n}} \right)^{n - 3} + \cdots } \right). $$ I put $n=1/x$ in the expression inside the parentheses and expanded about $x=0$ using computer algebra system. Then I replaced $x$ by $1/n$. With careful analysis of remainders, this could be made rigorous. – Gary Jun 29 '21 at 18:18
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    It may be shown that the coefficient of $\frac{1}{n^m}$ in my expansion will be $$ \frac{1}{{m!}}\left[ {\frac{{d^m }}{{dx^m }}\left( {\sum\limits_{k = 0}^m {x^k e^{ - k} \exp \left( {\sum\limits_{j = 1}^{m - k} {\frac{{k^{j + 1} }}{{j(j + 1)}}x^j } } \right)} } \right)} \right]_{x = 0} . $$ – Gary Jun 29 '21 at 18:25
  • @Gary Thank you so much for your effort!! Appreciate it!! Takes quite a long time for me to understand as I am new to series :) – Maximus Su Jun 29 '21 at 18:29
  • @Gary How do you get the equation after you said ''Starting from the "end" of the series,'' ? – Maximus Su Jun 30 '21 at 05:00
  • @Gary Is it by computer ? – Maximus Su Jun 30 '21 at 05:01
  • Just note that $$ 1^1 + 2^2 + \cdots + (n - 1)^{n - 1} + n^n = n^n + (n - 1)^{n - 1} + \cdots \ = n^n \left( {1 + \frac{{(n - 1)^{n - 1} }}{{n^n }} + \cdots } \right) = n^n \left( {1 + \frac{1}{n}\left( {\frac{{n - 1}}{n}} \right)^n + \cdots } \right). $$ – Gary Jun 30 '21 at 05:38

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