0

Series

Hi readers, may i know how this formula is derived? I thought of using https://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula which is Euler-Maclaurian to find the formula but its not.

Maximus Su
  • 51
  • As far as I know there is no nice closed form for the given sum. – Peter Jun 30 '21 at 05:26
  • 1
    Hi @Peter do you know how to derive the above expression? – Maximus Su Jun 30 '21 at 05:27
  • Isn't deriving the summation just grouping terms with the same exponent? – Smriti Sivakumar Jun 30 '21 at 05:28
  • Hi @Smriti Sivakumar If you know the answer, you can post it below so that it benefits all readers. – Maximus Su Jun 30 '21 at 05:30
  • You can also write it on a piece of paper and upload it – Maximus Su Jun 30 '21 at 05:31
  • $n^n$ has been "factored out" of every summand. The expression in the parantheses can be seen to be smaller than $1+1/n+1/n^2+1/n^3+...$ hence for large $n$, we get a good approximation by truncating at the $1/n^2$-term. Of course the absolute error is large since we multiply with $n^n$. – Peter Jun 30 '21 at 05:35
  • I’m voting to close this question because This could have been discussed here: https://math.stackexchange.com/questions/4186301/what-is-an-approximate-closed-form-for-sum-of-nn-series – Gary Jun 30 '21 at 05:41

1 Answers1

0

How I think the summation was derived:$$\sum_{k=1}^n k^k= 1^1 + 2^2 + 3^3+....+n^n$$ $$=n^n\left(1+ \frac{(n-1)^{n-1}}{n^n}+\frac{(n-2)^{n-2}}{n^n}+\frac{(n-3)^{n-3}}{n^n}+...+\frac{1^{1}}{n^n}\right)$$ $$=n^n\left(1+ \frac{(n-1)^{n-1}}{n^{n-1}\cdot n}+\frac{(n-2)^{n-2}}{n^{n-2}\cdot n^2}+\frac{(n-3)^{n-3}}{n^{n-3}\cdot n^3}+...+\frac{1^{1}}{n^1\cdot n^{n-1}}\right)$$ $$=n^n\left(1+ \frac{1}{n}\left(\frac{n-1}{n}\right)^{n-1}+\frac{1}{n^2}\left(\frac{n-2}{n}\right)^{n-2}+\frac{1}{n^3}\left(\frac{n-3}{n}\right)^{n-3}+...+\frac{1}{n^{n-1}}\left(\frac{1}{n}\right)^{1}\right)$$

Smriti Sivakumar
  • 999
  • Thank you Smriti, LOL, i thought this is a closed form but its just simplifying it. – Maximus Su Jun 30 '21 at 05:34
  • @MaximusSu haha exactly, i was confused about the question here, hence the "I think" – Smriti Sivakumar Jun 30 '21 at 05:34
  • so far there is no nice closed form for this right? There is a nice closed form of n^1.5 https://math.stackexchange.com/questions/1393811/sum-of-1-5-powers-of-natural-numbers – Maximus Su Jun 30 '21 at 05:36
  • @MaximusSu to my knowledge, no. – Smriti Sivakumar Jun 30 '21 at 05:37
  • But, its not accurate for n^n, because the number of terms required to cancel the error in Euler-Maclaurian series is just too many. – Maximus Su Jun 30 '21 at 05:37
  • @MaximusSu I discussed how to derive the asymptotics here: https://math.stackexchange.com/questions/4186301/what-is-an-approximate-closed-form-for-sum-of-nn-series Why did you open a new question? – Gary Jun 30 '21 at 05:42