# Convergence of this sum

• May 13th 2010, 08:48 PM
Bacterius
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 :)
• May 13th 2010, 09:02 PM
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Quote:

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 $\displaystyle \int_0^1 \frac{1}{x^x} \, dx$."
• May 13th 2010, 09:09 PM
Drexel28
Quote:

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, $\displaystyle \int_0^1 \frac{dk}{k^k}=\sum_{k=1}^{\infty}\frac{1}{k^k}$
• May 13th 2010, 09:19 PM
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 !
• May 14th 2010, 01:03 AM
CaptainBlack
Quote:

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
• May 14th 2010, 01:41 AM
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. $\displaystyle R_N < \frac{1}{N^N}$ )
• May 14th 2010, 02:25 AM
CaptainBlack
Quote:

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
• May 17th 2010, 03:15 AM
Bacterius
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 :o
• May 24th 2010, 06:46 PM
chiph588@
Quote:

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 (Giggle).

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.
• May 24th 2010, 06:52 PM
Drexel28
Quote:

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 (Giggle).

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 $\displaystyle \int_0^1 f(x)dx=\sum_{n=1}^{\infty}f(n)$?
• Aug 24th 2010, 05:43 PM
TerenceCS
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.(Doh)