Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In this article, the method of computing $1+2+\dots$ is outlined. Is there a similar method for computing $1^2+2^2+\dots$? What about for the general power $n$? (That is, $1^n+2^n+\dots$)

share|improve this question
1  
This is homework? –  Pedro Tamaroff Jul 26 '12 at 21:56
    
This question may prove useful –  MJD Jul 27 '12 at 2:32
    
check this wolframalpha article –  user31280 Oct 27 '12 at 4:42

2 Answers 2

To go to infinity you will need the Riemann zeta function.

For powers $n$ the result is given by $\zeta(-n)=-\frac{B_{n+1}}{n+1}$.
(this means $0$ for $n$ even since $B_n$ is a Bernoulli number)

The story is that $\displaystyle f(s)=\sum_{k=1}^\infty \frac 1{k^s}$ was found convergent for $s>1$ (Euler solved the famous 'Basel' case $s=2$) so that a function was searched later that worked for any complex $s\not= 1$ (Riemann).

For negative integer values of $s$ you'll get your limits.
In the wikipedia section kindly linked by Will Jagy note the Bernoulli numbers : that's what you really need! (Euler's teacher was Jean Bernoulli by the way...).

For the story of Euler's contribution see Sandifer's 'How Euler did it'.

Ramanujan rediscovered your series too but that's another story.

share|improve this answer

There is no similar method to compute $1^2+2^2+\cdots$. However, there is a trick for you to compute any finite term series sum like $\sum\limits_{k=0}^{m}k^n$.

To find such a sum, you first need to know the sums $\sum\limits_{k=0}^{m}{k^p}, \forall p<n$. Then you can use the following trick to find $\sum\limits_{k=0}^{m}k^n$.

  1. Rewrite $$\sum\limits_{k=0}^{m}k^{n+1}=\sum\limits_{k=0}^{m+1}k^{n+1}-(m+1)^{n+1}$$
  2. Rewrite the first term $$\sum\limits_{k=0}^{m+1}k^{n+1}=\sum\limits_{k=0}^{0}0^{n+1}+\sum\limits_{k=1}^{m+1}k^{n+1}=\sum\limits_{k=0}^{m}(1+k)^{n+1}$$
  3. Expand $(1+k)^{n+1}$ using binomial coefficients, i.e. $$(1+k)^{n+1}=\sum\limits_{p=0}^{{n+1}}{{n+1}\choose p}k^p1^{{n+1}-p}=\sum\limits_{p=0}^{n+1}{{n+1}\choose p}k^p=k^{n+1}+\sum\limits_{p=0}^{n}{{n+1}\choose p}k^p$$
  4. Put these pieces together, then you have $$\sum\limits_{k=0}^{m}k^{n+1}=\sum\limits_{k=0}^{m+1}k^{n+1}-(m+1)^{n+1}=\sum\limits_{k=0}^{m}\left(k^{n+1}+\sum\limits_{q=0}^{n}{{n+1}\choose p}k^p\right)-(m+1)^{n+1}\\=\sum\limits_{k=0}^{m}k^{n+1}+\sum\limits_{q=0}^{n}{{n+1}\choose p}\left(\sum\limits_{k=0}^{m}k^p\right)-(m+1)^{n+1}$$ implying that$$\sum\limits_{p=0}^{n}{{n+1}\choose p}\left(\sum\limits_{k=0}^{m}k^p\right)-(m+1)^{n+1}=0$$ Therefore, we have $${{n+1}\choose n}\sum\limits_{k=0}^{m}k^n=(m+1)^{n+1}-\sum\limits_{p=0}^{n-1}{{n+1}\choose p}\left(\sum\limits_{k=0}^{m}k^p\right)$$ namely $$\sum\limits_{k=0}^{m}k^n=\frac{1}{n+1}\left((m+1)^{n+1}-{{n+1}\choose n-1}\sum\limits_{k=0}^{m}k^{n-1}-\cdots-{{n+1}\choose 1}\sum\limits_{k=0}^{m}k-{{n+1}\choose 0}\sum\limits_{k=0}^{m}1\right)$$ Since you know for $n=1$$$\sum\limits_{k=0}^{m}k=\frac{m(m+1)}{2}$$, then use the above trick you can derive $\displaystyle \sum\limits_{k=0}^{m}k^2$, $\displaystyle \sum\limits_{k=0}^{m}k^3\cdots$,$\displaystyle \sum\limits_{k=0}^{m}k^{n-1}$, and finally $\displaystyle \sum\limits_{k=0}^{m}k^n$.
share|improve this answer
    
This is, sadly, not what the OP is asking. –  Pedro Tamaroff Jul 27 '12 at 3:12

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.