General rule for a confusing series

[Jesseac]

Hi,

I am a year 12 IB maths HL student... and i was wondering about the possibility of a general rule for a particular problem... I was looking at the following rules, which came up in a basic textbook exercise on mathematical induction.

1+2+3+4+...+n=(n^2+n)/2 where n is a positive integer
(1^2)+(2^2)+...+(n^2)=(n(n+1)(2n+1))/6, where n is a positive integer

I was wondering if there exists a rule for the case.

(1^a)+(2^a)+(3^a)+(4^a)+...+(n^a), where both n and a are positive integers.

This problem has been bugging me for quite some time. I had the feeling that the problem may be slightly different for odd values of a as opposed to even values of a.

Also, if there does indeed exist a general rule or rules for this case, I was wondering if someone could give an outline as to the method used to prove this (as I'm not so confident in attempting it for myself).

 

[zetafunction]

yes it exists in terms of the Hurwitz zeta function or Bernoulli Polynomials

if 'm' is an integer

$$1+2^{m}+...+k^{m} = \frac{B_{m+1}(k)-B_{m+1}(0)}{m+1}$$

otherwise you need the Hurwitz zeta function.. look at this at wikipedia.

 

[Stephen Tashi]

if someone could give an outline as to the method used to prove this (as I'm not so confident in attempting it for myself).

I don't know a general rule for discovering formulas for sums, but proving they work is often straightforward. If you think that $$\sum_{i=0}^n a_i = G(n)$$
then see if you can write $$G(n)$$ as $$F(n+1) - F(0)$$ for some $$F(n)$$ with the property that $$\triangle F(n) = F(n+1) - F(n) = a_n$$.

If you can do that, you're done, since for such a $$F(n)$$,

$$\sum_{i=0}^n a_n = \sum_{i=0}^n (F(n+1)-F(n))$$

$$=(F[1] - F[0]) + (F[2] - F[1] ) + (F(3) - F(2)) + ... (F(n+1) - F(n) )$$

which "telescopes" to $$F(n+1) - F(0)$$

Example: $$a_n = n$$
$$G(n) = \frac{(n)(n+1)}{2}$$
$$F(n) = \frac{(n-1)(n)}{2}$$

$$\triangle F(n) = \frac{(n)(n+1)}{2} - \frac{(n-1)(n)}{2}$$
$$= \frac{n^2 + n}{2} - \frac{n^2 - n }{2} = n = a_n$$
You can also invent summation formulas for complicated looking sequences by setting $$F(n)$$ equal to some formula and working backwards to the series that it sums.