Linked Questions

3 votes
1 answer
1k views

Sum of self power [duplicate]

Is there a formula to calculate the sum of a number to the power of this same number, like: $$1^1 + 2^2 + 3^3 + 4^4 + 5^5 + ... + n^n$$? or $$x^x + (x+1)^{(x+1)} + (x+2)^{(x+2)} + ... + (x+n)^{(x+...
PPP's user avatar
  • 2,041
10 votes
2 answers
703 views

Is there any general formula for $S = 1^1 + 2^2 + 3^3 + \dotsb+(n - 1)^{n - 1} + n^n, n \in N$? [duplicate]

Is there any general formula to sum following series: $$S = 1^1 + 2^2 + 3^3 + \dotsb+(n - 1)^{n - 1} + n^n, n \in N$$ I mean for $S = f(n)$, is there a formula to compute $f(n)$?
Vishal's user avatar
  • 321
1 vote
1 answer
192 views

Formula to calculate sum of $k^k $ upto $n$ terms. [duplicate]

What will be the general formula for the series: $1^1 + 2^2 + 3^3 + 4^4\ldots$ The series diverges, but can we have a formula to find the sum of $n$ terms.
Harsha G.'s user avatar
  • 412
3 votes
0 answers
119 views

How to find $\sum_{k=1}^n k^k$? [duplicate]

Actually question which I found: Find the sum of the series $1^1+2^2+3^3+ \cdots +n^n $ This question has been bothering me since a long time. Any help would be appreciated!
Harshal Gajjar's user avatar
4 votes
0 answers
71 views

Is there a formula to calculate the following sum? [duplicate]

I tried googling, but I may not be searching with correct terms. $$\sum_{k=1}^{n} k^k = 1^1 + 2^2 +3^3 + 4^4 + \cdots + n^n$$ So, is there a formula to calculate this?
Durmus's user avatar
  • 181
1 vote
0 answers
89 views

How to find out the sum of the series $1^1+2^2 +\ldots+n^n?$ [duplicate]

Recently, I was exploring different ways to find out the sum of series but I've one in my mind to which I didn't have a sol. to. It goes :$$\sum_{i=1}^n i^i$$ Does anyone have a solution to this ?
user avatar
2 votes
0 answers
67 views

Closed form for $\sum_{n=1}^{m} n^n$ [duplicate]

I would like to ask if anyone would help me to find a general formula for the sum of m members of the following series. \begin{equation} \sum_{n=1}^{m} n^n = 1^1+2^2+\cdots + m^m \end{equation} ...
user avatar
1 vote
0 answers
41 views

A partial sum formula [duplicate]

I'm very familiar with partial sums and such little bit hard once, but I was wondering is there a partial sum formula for that $$\displaystyle\sum_{n=1}^k n^n$$ I have tried with Wolfram alpha but I ...
I'mmksk's user avatar
  • 21
4 votes
3 answers
171 views

$\lim_{n\to\infty}{\left({\sum_{k=1}^nk^k}\right)/{n^n}}=1$?

I'm interested in the following sum $S_n$. $$S_n:=\sum_{k=1}^nk^k=1^1+2^2+3^3+\cdots+n^n.$$ Letting $T_n:={S_n}/{n^n}$, wolfram tells us the followings. $$T_5=1.09216, T_{10}\approx1.04051, T_{30}\...
mathlove's user avatar
  • 145k
1 vote
1 answer
123 views

Does it exist a recursive formula that approximates $\sum_{k=1}^n k^k\,$?

Does it exist a recursive formula that approximates $\sum_{k=1}^n k^k\,$? Defined $\,S_n=\sum_{k=1}^n k^k$, I am looking for a relation $\,S_{n+1}\sim f(n)\cdot S_n\,$, with $\,S_1=1$ and $f(n)\,$ a ...
Augusto Santi's user avatar
2 votes
0 answers
142 views

Why there isn't any formula to calculate $\sum\limits_{k=1}^n{k^k}$ explicitly?

I've looked over this Closed form formula for $\sum\limits_{k=1}^n k^k$ but it doesn't have any explicit formula, only the bounds. So, is it the case that no general formula can exist to calculate ...
ankit's user avatar
  • 2,411