Merry Christmas /sci/, could any of you gentlemen tell me if the following series converges to any number? And if so what number?
Does not converge
>>7743397
okay, so if I get this right, it doesn't converge because x^x grows faster than then the range can decline?
>>7743400
Lol i thought its a lambda
Start writing your homework proberly if you want hell faggot
here you go
>>7743402
It's not homework, no one does homework on Christmas dude
>>7743394
I can tell you that it converges (the limit of, the integral of x^x between x=1/x and zero, as x approaches infinity is equal to zero) but I don't know how to find the number to which that series converges.
inb4 lrn 2 latex
>>7743534
lol wtf am I saying, it diverges. 1/infinity =/= 0 for convergence tests.
>>7743412
[math]
\displaystyle \sum _x ^\infty \int _0 ^\frac{1}{x} x^x dx
[/math]
>>7743412
>hij gebruikt geen LaTeX
>>7743534
That the sequence converges to zero is a necessary but not sufficient condition for the convergence of the sequence.
Also, can someone explain the summation notation that this problem is using? What are you summing over?
>>7743574
*convergence of the series.
My bad.
>>7743574
from x = 0 to x = ∞
>>7743413
I have to do 200 different integrals for homework (200 variations of x^n does not count).
>>7743412
Is that Dutch?, looks like German but with a retardation factor of 2.5 Schwarzeneggers.
In the interval [math][0,1][/math], the function [math] x^x [/math] has a single minimum at
[math] x = \frac {1} {e} [/math],
see
https://en.wikipedia.org/wiki/Secretary_problem
and it's [math] ( \frac {1} {e} )^{ \frac {1} {e} } > \frac {1} {2} [/math].
Thus
[math] \forall (x \in [0,1]) \ \ (2 · x^x) > 1[/math]
Thus
[math] \int_0^{ r } 2·x^x \, dx > \int_0^{ r } dx > r [/math]
and in particular
[math] \int_0^{ \frac {1} {n} } 2·x^x \, dx \, > \, \frac {1} {n} [/math]
and of course the integral and the sum is linear (so you can push in a factor of 2 under the integrand) and then the whole thing is
[math] > \sum_{n=1}^\infty \frac {1} {n} = \infty [/math]
>>7743600
then your first term is not well defined
>>7743751
isnt that like 2, and not infinity?
>>7743742
Good solution
>>7743755
HAHAHAHAHAHA HAHAHA
>>7743755
No, but the sum of the inverse is somehow -1/12.
>>7743649
Underrated post
wrong proof
>>7743394
Fairly sure OP means to inquire about the following series.
[math]
\sum\limits_{x}^{\infty} \int_0^{1/x} t^t dt
[\math]
>>7743751
I didnt even figure it like that and got inf -1 = inf
>>7743394
how do you index over x? do you assume x is a positive integer?
>>7743412
kankermongool
