# Partial sum of divergent series

I am trying to find the nth partial sum of this series: $$S(n) = 2(n+1)^2$$

I found the answer on WolframAlpha:

$$\sum_{n=0}^m (1+2n)^2 =\frac{1}{3}(m+1)(2m+1)(2m+3)$$

How can I calculate that sum, without any software?

• It's a sum of squares. You may learn from math.stackexchange.com/q/188602/290189 – GNUSupporter 8964民主女神 地下教會 Oct 1 '18 at 20:41
• $2(n+1)^2\ne(1+2n)^2$. Which one are you trying to calculate? – Andrei Oct 1 '18 at 20:43
• Hint: \begin{eqnarray*} \sum_{n=0}^{m} n^2=\frac{m(m+1)(2m+1)}{6} \end{eqnarray*} – Donald Splutterwit Oct 1 '18 at 20:44
• @Andrei Sorry, that is a mistake. I am trying to calculate $(1+2n)^2$ – GKEdv Oct 1 '18 at 20:48

$$S(n)=(1+2n)^2=1+4n+4n^2$$ You can now use the following $$\sum_{n=0}^m1=m+1\\\sum_{n=0}^mn=\frac{m(m+1)}{2}\\\sum_{n=0}^mn^2=\frac{m(m+1)(2m+1)}{6}$$

Alternatively, compute the first 4-5 elements. The sum of a polynomial of order $$p$$ will be a polynomial of order $$p+1$$ in the number of terms. Find the coefficients, then prove by induction

• Thanks, the first approach is very easy to understand. Can you elaborate on the alternative approach or point me to more information, example on that? – GKEdv Oct 1 '18 at 21:20
• If you look at en.wikipedia.org/wiki/Faulhaber%27s_formula#Summae_Potestatum you can get that the sum of $n^p$ is a polynomial of order $p+1$. Adding then the rest of the smaller powers will change the coefficients, but not the order of the polynomial. – Andrei Oct 1 '18 at 21:44
• The first sum should be $m\color{red}{+1}$ ... it starts at zero! – Donald Splutterwit Oct 2 '18 at 1:39
• You are right. It does not matter for the other ones. Thanks. I'll fix it – Andrei Oct 2 '18 at 3:14

$$\sum_\limits{i=0}^n 2(i + 1)^2 = 2\sum_\limits{i=1}^{n+1} i^2$$

Which gets to the meat of the question, what is $$\sum_\limits{i=1}^n i^2$$?

There are a few ways to do this. I think that this one is intuitive. In the first triangle, the sum of $$i^{th}$$ row equals $$i^2$$

The next two triangles are identical to the first but rotated 120 degrees in each direction.

Adding corresponding entries we get a triangle with $$2n+1$$ in every entry. What is the $$n^{th}$$ triangular number?

$$3\sum_\limits{i=1}^n i^2 = (2n+1)\frac {n(n+1)}{2}\\ \sum_\limits{i=1}^n i^2 = \frac {n(n+1)(2n+1)}{6}$$

To find: $$\sum_\limits{i=1}^{n+1} i^2$$, sub $$n+1$$ in for $$n$$ in the formula above.

$$\sum_\limits{i=0}^n 2(i + 1)^2 = \frac {(n+1)(n+2)(2n+3)}{3}$$

Another approach is to assume that $$S_n$$ can be expressed as a degree $$3$$ polynomial. This should seem plausible

$$S(n) = a_0 + a_1 n + a_2 n^2 + a_3n^3\\ S(n+1) = S(n) + 2(n+2)^2\\ S(n+1) - S_n = 2(n+2)^2\\ S(n+1) = a_0 + a_1 (n+1) + a_2 (n+1)^2 + a_3(n+1)^3\\ a_0 + a_1 n+a_1 + a_2 n^2 + 2a_2n+a_21 + a_3n^3 + 3a_3n^2 + 3a_3n + 1\\ S(n+1) - S(n) = (a_1 + a_2 + a_3) + (2a_2 + 3a_3) n + 3a_3 n^2 = 2n^2 + 4n + 2$$

giving a system of equations:

$$a_1 + a_2 + a_3 = 2\\ 2a_2 + 3a_3 = 4\\ 3a_3 = 1\\ a_0 = S(0)$$