Two weeks ago, at Metroplex Math Circle presentation "Primes and Composite Numbers", the presenter used a formula for the sum of the squares of numbers from 1 to n in his solution of a problem. I thought that I might be able to derive the formula.

1^2+2^2+ .......+n^2= n*(n+1)*(2n+1)/6

To derive the formula turns out to be not trivial, but to prove it was not hard, using induction. During my attempt to derive the formula, I tried converting the problem into a recursive procedure, into a continuous problem instead of discrete problem so that I could use calculus....., all without success (1). The idea of using continuous function for solving discrete number problem stays with me.1^2+2^2+ .......+n^2= n*(n+1)*(2n+1)/6

The next day, I encountered another formula for the calculation of Pi

Pi = 4 ( 1/1 - 1/3+1/5-1/7+1/9-1/11+1/13 ......+1/(2n-1) - 1/(2n+1) + ...........)

The irrational Pi can be determined by this infinite series of rational numbers - fractions!!

**The sum of infinite numbers**

My typical approach to find the sum of infinite numbers is to find a formula for the sum of first n terms of the series, and then take a limit (n--> infinite). For example

1/(1*2) + 1/(2*3)+1/(3*4) + ......+ 1/(n*(n+1))+ ......

Let S be the sum of the infinite series, Sn be the sum of the first n terms

Note that 1/(n*(n+1)) = 1/n - 1/(n+1), so

Sn = 1/1 -1/2+1/2-1/3 +1/3-1/4 + ...1/n -1/(n+1) = 1-1/(n+1)

S= Sn(N--> infinite) = 1

This approach is apparently not suitable for the calculation of the formula for Pi listed above

**Taylor Expansion for sum of infinite series**

Looking at the formula

Pi = 4 ( 1/1 - 1/3+1/5-1/7+1/9-1/11+1/13 ......+1/(2n-1) - 1/(2n+1) + ...........)

the idea of using continuous function to calculate the discrete came back to me. This formula seems to relate to trigonometric functions, arcsin or arctan. Checking handbook of math, I found this formulation of Taylor expansion of arctan (x) when -1 <= x <=1)

arctan(x) = x- x^3/3+x^5/5-x^7/7 +x^9/9 .....+(-1)^n x^(2n+1)/(2n+1) + .....

Take x=1, the above Taylor expansion yield

Pi = 4 ( 1/1 - 1/3+1/5-1/7+1/9-1/11+1/13 ......+1/(2n-1) - 1/(2n+1) + ...........)

With Taylor Expansion of various functions, one can easily determine many seemly impossible sum of infinite series .

(1) Upon literature search, it turned out that more general solution exists, and the sum of powers (fixed exponent m) is always a polynomial of n of degree m+1. I will find the derivation of it and summarize it in a future post.

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