Is n^n+1 a Valid Formula for the Sum of n^n Series?

[mxam]

Can you help me with this exercise?

1^{1}+2^{2}+3^{3}+4^{4}+...+n^{n} = n^{n+1}

Thanks!

PD. I was trying to solve, and i have this:

1^{1}=1^{1+1} =
1 = 1

a) k^{k+1}
b) k+1^{k+1} = k+1^{(k+1)+1}

a in b) k^{k+1} + k+1^{k+1} = k+1^{(k+1)+1}

(k)^{k}(k)^{1}+(k+1)^{k}(k+1)^{1}=(k+1)^{k}(k+1)^{1}(k+1)^{1}

I´m lost in this step . . . Thanks again!

 

[dextercioby]

Where did you pick the exercise from? It looks wrong.

 

[mxam]

The Algebra Teacher gives the exercise, but i don't know, maybe don´t have a solution . . . ?

 

[jedishrfu]

The series and its sum don't seem correct.

Can you check your book again?

1+2^2 =/= 2^(2+1)

1+4 = 5 but 2^(2+1) = 8

 

[jedishrfu]

There are many series that look vaguely similar like

1 + 2 + 3 + ... = (n*(n+1))/2

 

[mxam]

The series and its sum don't seem correct.

Can you check your book again?

1+2^2 =/= 2^(2+1)

1+4 = 5 but 2^(2+1) = 8

A attach the picture of the book. . .

 

[jedishrfu]

Can you say there is no solution and prove that by example?

 

[mxam]

Can you say there is no solution and prove that by example?

The teacher said that exist a solution, but with the opinions and the examples i think that he´s lying. . . =o!

 

[jedishrfu]

The teacher said that exist a solution, but with the opinions and the examples i think that he´s lying. . . =o!

Maybe strategically misdirecting you in order to have you discover this alternative...

You could PM a PF Mentor to see if they can add some new insight...

 

[gopher_p]

I am curious. What are the instructions for the exercise? You did not include them in the original post, and your picture does not show them either.

 

[mxam]

I am curious. What are the instructions for the exercise? You did not include them in the original post, and your picture does not show them either.

Check the Attachment, the exercise says: "Demuestre por Induccion Matematica las siguientes preposiciones dadas" or in English, "Demonstrate by Mathematical Induction the following prepositions" . . . I think is a exercise that don't have a real solution, maybe the teacher´s proving us, . . . But I´ll need a good explanation for this, =)

 

[vanhees71]

The formula is obviously wrong.

Much more interesting is the question, how to figure out such sums. One clever trick is to use an easier sum to derive the one you want to know. Here the obvious idea is to use the generating function
g(x)=\sum_{k=0}^n \exp(k x).
This is a geometric series. You should prove, e.g., by induction, that it gives
g(x)=\frac{\exp[(n+1) x)-1}{\exp x-1}.
Then you can get other sums by taking derivatives of this, e.g.,
g'(x)=\sum_{k=0}^{n} k \exp(k x) \; \Rightarrow \; g'(0)=\sum_{k=0}^n k,
and so on. There's, of course a subtlety with taking x=0, which you also should discuss.

 

[HallsofIvy]

Can you help me with this exercise?

1^{1}+2^{2}+3^{3}+4^{4}+...+n^{n} = n^{n+1}

Thanks!

PD. I was trying to solve, and i have this:

1^{1}=1^{1+1} =
1 = 1

Yes, this is true for n= 1. But surely you saw that when n= 2, the left side is 1^1+ 2^2= 1+ 4= 5 while the right side is 2^3= 8.

a) k^{k+1}
b) k+1^{k+1} = k+1^{(k+1)+1}

a in b) k^{k+1} + k+1^{k+1} = k+1^{(k+1)+1}

(k)^{k}(k)^{1}+(k+1)^{k}(k+1)^{1}=(k+1)^{k}(k+1)^{1}(k+1)^{1}

I´m lost in this step . . . Thanks again!

 

[mxam]

I´ve 1 curious question:

If i want to know what´s the correct serie for n^{n+1}, what´s the the procedure to follow?

 

[dirk_mec1]

Use a Taylor series.

 

[jedishrfu]

Doing the first nine iterations:

n=1:	1		        sum/psum = 1
n=2:	5		        sum/psum = 5
n=3:	32		        sum/psum = 6
n=4:	288		        sum/psum = 9
n=5:	3413		        sum/psum = 11
n=6:	50069		        sum/psum = 14
n=7:	873612		        sum/psum = 17
n=8:	17650828		sum/psum = 20
n=9:	405071317		sum/psum = 22

Processing IDE code used:

void setup() {
  
 int nterms = 10;
 int psum=1, sum=0;
 
 for(int n=1; n<nterms; n++) {
    sum+=Math.pow(n,n);
    int iratio = sum/psum;
    println("n="+n+":\t"+sum+"\t\tsum/psum = "+iratio);
    psum=sum;
 }
 
}