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# Some thoughts about the sums

On this note, I said about some patterns, even though the Faulhaber's formula makes this somewhat useless, I found some patters, first of all, as someone commented:

$$\sum^n_{i=1}i^a\approx\frac{n^{a+1}}{a+1}$$

I will prove the following statement:

"The highest order term of the polynomial that represents $$\sum^n_{i=1}i^a$$ is $$\frac{n^{a+1}}{a+1}$$"

I will use a strong induction to prove this, starting with the base case of $$a=0$$, we have:

$$\displaystyle \sum^n_{i=1}i^a=n$$ So the highest term is $$n=\frac{n^{a+1}}{a+1}$$

We will assume that this will work for $$a=0,1,2,\ldots,k$$, then for $$a=k+1$$, we have:

$$\displaystyle \sum^n_{j=1}j^{k+1}=\frac{\sum^{k+2}_{i=1}\binom{k+2}{i}n^i-\sum^k_{i=0}\left (\binom{k+2}{i}\sum^n_{j=1}j^i \right )}{k+2}$$

The highest term of this will be $$\frac{n^{k+2}}{k+2}$$, that is obtained when, on the first sum, the $$i=k+2$$, so $$\binom{k+2}{i}n^i=\binom{k+2}{k+2}n^{k+2}=n^{k+2}$$, no other term has this order on the numerator, because the highest one on the other sum is $$\frac{n^{k+1}}{k+1}$$, whose order is not the same, the term becomes $$\frac{n^{k+2}}{k+2}$$ when I consider the denominator;

The other pattern I saw is that the second term of the sum is $$\frac{n^a}{2}$$

I will have to use the Faulhaber's formula:

$$\sum^n_{i=1}i^a=\frac{1}{a+1}\sum^a_{i=0}\binom{a+1}{i}B^+_in^{a+1-i}$$

The term of $$a$$ degree appears when i=1 on the right sum, so it is:

$$\frac{1}{a+1}\binom{a+1}{1}B^+_1n^a=B^+_1n^a$$

Since $$B^+_1=\frac{1}{2}$$ (I am using this definition), the term is:

$$\frac{1}{2}n^a=\frac{n^a}{2}$$

Note by Matheus Jahnke
9 months, 3 weeks ago

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These are great thoughts! Mathematicians look for patterns in the data, and then try to prove that the pattern exists, as you have done above.

Staff - 9 months, 3 weeks ago