1. Convergence of this sum

Hello,
In chemistry class, instead of working hard on valence shells and bonds, I've been fiddling with my calculator. I tried to evaluate the following sum out of interest :

$\displaystyle \sum_{k = 1}^{\infty} \frac{1}{k^k} \approx 1.291285 \cdots$

It converges. Has anyone ever seen this actual sum somewhere ? Does it have a particular meaning or use, as $\displaystyle \zeta{(s)}$ ? Or is it just yet another random and useless summation ? I'm asking the question because I feel this sum has something special, having the variable both in base and exponent.

I've been looking on Google but didn't find any reading on it. Any input is appreciated

2. Originally Posted by Bacterius
Hello,
In chemistry class, instead of working hard on valence shells and bonds, I've been fiddling with my calculator. I tried to evaluate the following sum out of interest :

$\displaystyle \sum_{k = 1}^{\infty} \frac{1}{k^k} \approx 1.291285 \cdots$

It converges. Has anyone ever seen this actual sum somewhere ? Does it have a particular meaning or use, as $\displaystyle \zeta{(s)}$ ? Or is it just yet another random and useless summation ? I'm asking the question because I feel this sum has something special, having the variable both in base and exponent.

I've been looking on Google but didn't find any reading on it. Any input is appreciated
Interesting! The best I could come up with required no original thought and came from OEIS, sequence A073009:

"This is also equal to $\displaystyle \int_0^1 \frac{1}{x^x} \, dx$."

3. Originally Posted by Bacterius
Hello,
In chemistry class, instead of working hard on valence shells and bonds, I've been fiddling with my calculator. I tried to evaluate the following sum out of interest :

$\displaystyle \sum_{k = 1}^{\infty} \frac{1}{k^k} \approx 1.291285 \cdots$

It converges. Has anyone ever seen this actual sum somewhere ? Does it have a particular meaning or use, as $\displaystyle \zeta{(s)}$ ? Or is it just yet another random and useless summation ? I'm asking the question because I feel this sum has something special, having the variable both in base and exponent.

I've been looking on Google but didn't find any reading on it. Any input is appreciated
It has an interesting property

$\displaystyle \int_0^1 k^kdk=\int_0^1 e^{k\ln(k)}dk=\int_0^1\sum_{n=0}^{\infty}\frac{k^n \ln^n(k)}{n!}=\sum_{n=0}^{\infty}\frac{1}{n!}\int_ 0^1 k^n\ln^n(k)$.

But, now for the integral let $\displaystyle z=-\ln(k)\implies k=e^{-z}\implies dk=-e^{-z}dz$ and so our integral becomes $\displaystyle -\int_{\infty}^0 e^{-(n+1)z}z^n dz=\int_0^{\infty} z^n e^{-nz}$. So, let $\displaystyle (n+1)z=\xi$ and this becomes $\displaystyle \frac{1}{(n+1)^{n+1}}\int_0^{\infty}\xi^n e^{-z}d\xi=\frac{n!}{(n+1)^{n+1}}$ and so our sum is equal to $\displaystyle \sum_{n=0}^{\infty}\frac{1}{(n+1)^{n+1}}=\sum_{n=1 }^{\infty}\frac{1}{n^n}$

Thus, $\displaystyle \int_0^1 \frac{dk}{k^k}=\sum_{k=1}^{\infty}\frac{1}{k^k}$

4. Although I am not a bookworm , all of the books i have read never mention how to find the sum in an analytical way, at most they only explain the relation with the integral . I think the most reasonable way to find the sum is using the calculator because the series converges so fast !

5. Originally Posted by simplependulum
Although I am not a bookworm , all of the books i have read never mention how to find the sum in an analytical way, at most they only explain the relation with the integral . I think the most reasonable way to find the sum is using the calculator because the series converges so fast !
Im sure I posted this earlier, but it seems to have disappeared so here it is again.

This raises the question of the remainder $\displaystyle R_N$ when the sum is truncated after $\displaystyle N$ terms:

$\displaystyle \sum_{k=1}^{\infty}\frac{1}{k^k}=\sum_{k=1}^{N}\fr ac{1}{k^k}+R_N$

I can show that $\displaystyle 0<R_N<\frac{1}{(N-1)N^{N-1}}$ , but I expect someone can do better (there is plenty of elbow room for improvement for largish $\displaystyle N$).

CB

6. I am not sure ... please check it

$\displaystyle R_N = \frac{1}{(N+1)^{N+1}} + \frac{1}{(N+2)^{N+2}} + ...$

I consider $\displaystyle f(x) = \frac{1}{x^x}$ and let

$\displaystyle A$ be the sum of the area of the rectangles with dimensions $\displaystyle 1 \times f(N+1) , 1 \times f(N+2) , ....$ , $\displaystyle A = R_N$

which is smaller than the curved area under $\displaystyle f(x)$ since $\displaystyle f(x)$ is decreasing .

Therefore , $\displaystyle R_N = A < \int_N^{\infty} f(x)~dx = \int_N^{\infty} \frac{dx}{x^x}$

$\displaystyle < \int_N^{\infty} \frac{1}{x^x} (1 + \ln{x} )~dx$

$\displaystyle = \left[ \frac{-1}{x^x} \right]_N^{\infty} = \frac{1}{N^N}$

If my calculations are all correct , it is interesting that the infinite sum starting from the next term is always smaller than that term . ( i.e. $\displaystyle R_N < \frac{1}{N^N}$ )

7. Originally Posted by simplependulum
I am not sure ... please check it

$\displaystyle R_N = \frac{1}{(N+1)^{N+1}} + \frac{1}{(N+2)^{N+2}} + ...$

I consider $\displaystyle f(x) = \frac{1}{x^x}$ and let

$\displaystyle A$ be the sum of the area of the rectangles with dimensions $\displaystyle 1 \times f(N+1) , 1 \times f(N+2) , ....$ , $\displaystyle A = R_N$

which is smaller than the curved area under $\displaystyle f(x)$ since $\displaystyle f(x)$ is decreasing .

Therefore , $\displaystyle R_N = A < \int_N^{\infty} f(x)~dx = \int_N^{\infty} \frac{dx}{x^x}$

$\displaystyle < \int_N^{\infty} \frac{1}{x^x} (1 + \ln{x} )~dx$

$\displaystyle = \left[ \frac{-1}{x^x} \right]_N^{\infty} = \frac{1}{N^N}$

If my calculations are all correct , it is interesting that the infinite sum starting from the next term is always smaller than that term . ( i.e. $\displaystyle R_N < \frac{1}{N^N}$ )
Not really that different from the earlier bound (their ratio goes to $\displaystyle 1$ and $\displaystyle N$ becomes absurdly large)

CB

8. Thanks for all the help
I appreciate it.
I'll have to investigate this relationship with the integral more deeply. There might be something to see

9. Originally Posted by Drexel28
$\displaystyle \int_0^1 \frac{1}{x^x}dx=\sum_{k=1}^{\infty}\frac{1}{k^k}$
This is called the "Sophmore's Dream". Only in a dream can $\displaystyle \int_0^1 f(x)dx=\sum_{k=1}^{\infty}f(k)$ (or some variation of that) always be true .

You may ask yourself what's the "Freshman's Dream"?. Well it's the fallacy $\displaystyle (a+b)^n = a^n+b^n \;\;\forall \;a,b,n$, a common mistake amongst some highschoolers.

10. Originally Posted by chiph588@
This is called the "Sophmore's Dream". Only in a dream can $\displaystyle \int_0^1 f(x)dx=\sum_{k=1}^{\infty}f(k)$ (or some variation of that) always be true .

You may ask yourself what's the "Freshman's Dream"?. Well it's the fallacy $\displaystyle (a+b)^n = a^n+b^n \;\;\forall \;a,b,n$, a common mistake amongst some highschoolers.
Interesting question, can you classify all $\displaystyle f\in C^{\infty}[0,1]$ (or some variant) such that $\displaystyle \int_0^1 f(x)dx=\sum_{n=1}^{\infty}f(n)$?

11. This series can be extended to a more general function:
$\displaystyle \displaystyle \omega(x)=\sum_{k=1}^{\infty} \frac{x^{k-1}}{k^k}=\int_{0}^{1} u^{-xu}\, du.$

You can check it and this function is define for all real $\displaystyle x$ and it's probably monotonically increasing (just found experimentally). It's non-negative for all real $\displaystyle x$.

Besides, straightly from the Taylor expansion of the function:
$\displaystyle \displaystyle \omega^{(n)} (0)=\frac{n!}{(n+1)^{n+1}}$

So
$\displaystyle \displaystyle \omega(1)=\sum_{k=1}^{\infty} \frac{1}{k^k}=\sum_{n=0}^{\infty} \frac{\omega^{(n)}(0)}{n!}.$
But this is just trivial.