The Sum n. Introduction. Proof I: By Induction. Caleb McWhorter

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1 The Sum n Caleb McWhorter Introduction Mathematicians study patterns, logic, and the relationships between objects. Though this definition is so vague it could be describing any field not just a scientific field! Where Mathematics differs from the others is the notion of a proof. But what exactly is that? A proof is one of those many nebulous words that refuse to be pinned down to a single concrete definition. But we will define a proof as an argument that uses assumed or known facts to establish a particular statement. Learning what a proof is and how to write them is an important part of a mathematician s early development. In fact, it is a lifelong process. One of the first proof techniques introduced to students is the Principle of Mathematical Induction. As an early example, students are shown the identity n i n() and asked to prove it using induction. 1 Though there are many ways of establishing this fact. Our goals here are twofold. First, we hope to demonstrate several different proofs of this fact using a variety of different (but related) techniques. Second, we hope to demonstrate that mathematical proofs can take many forms. Proofs of the same fact written by different people can look very different. What is important is that they convincingly (usually this is means rigorously whatever that means) establish the claim it endeavors to prove. After reading through the various proofs, the reader should look carefully to see what the various proofs have in common. Proof I: By Induction Proof. We proceed by induction. First, we check the first few cases by hand n 1 : n : n 3 : 1 i 1; i 1 + 3; 3 i ; 1(1 + 1) ( + 1) 3(3 + 1) This formula is a special case of Faulhaber s Formula. The values produced by the formula are called triangular numbers. 1

2 Now assume the result is true for n 1,, 3,..., k. We need to show that the formula holds for n k + 1. k+1 i (k + 1) + k i (k + 1) + k + k + k + k(k + 1) k + 3k + (k + 1) + k(k + 1) (k + 1)(k + ) k + + k + k (k + 1(k + 1) + 1) where the starred equality follows from the induction hypothesis: k i k(k+1). But then n i n() follows by induction. Proof II: By Dots Proof. Think of numbers as representing some number of dots, i.e. 1 represents one dot, represents dots, et cetera. Observe the sum represents the total number of dots in the diagram in Table 1. Table 1: The triangular array for n 5. We can represent the sum n as a total number of dots in diagram. The case for n 5 is demonstrated in Table 1. We can place two such diagrams next to each other. Observe we can remove the hypotenuse of the right triangular array, place it on top of the left triangular array and form a square, placing the removed hypotenuse s circles in a line to the right. Table : Adding the arrays for the case n 5. + The sum of the diagrams on the left is precisely the number of dots in the two triangular arrays: ( n) + ( n) ( n). The number of dots on the right of the equality is the area of the square, n n n, plus the number of dots in the column, n. Since both computations

3 calculate the number of dots in the diagram, we have ( n) n + n ( n) n(n + 1) n(n + 1) n But then n i n() Proof III: By Gauss Trick Proof. (Gauss ): Write out the sum in increasing order and again directly beneath it in decreasing order. Adding these two rows yields n n + (n 1) + (n ) n n + (n 1) + (n ) (n + 1) + (n + 1) + (n + 1) + + (n + 1) This result is the n-fold sum of (n + 1) s. But then we have ( n) n(n + 1) so that n i n n() Proof IV: By Summation Prestidigitation Proof. We want to find 1++ +n n i. Observe this is the same as finding n+(n 1)+ ++1 n n i + 1, the sum written in reverse. But then we have i i + n i + 1 i + (n i + 1) n + 1 n(n + 1) But then we have n i n() The story goes though there are many versions that when Gauss was in primary school, his teacher asked the class to practice addition by adding all the numbers from 1 to 100. Gauss came up with the correct sum of 5,050 in only a few seconds by thinking of the addition in the same way that we prove the theorem in Proof III. 3

4 Proof V: By More Summation Magic Proof. Let S (n) : n i0 i. Observe that S (n) S (n 1) n for n 1,,. But then S (n) is a polynomial of degree two. 3 Suppose that S (n) an + bn + c. We must have c 0 as S (0) 0. Furthermore, S (n) S (n 1) n and n S (n) S (n 1) (an + bn) (a(n 1) + b(n 1)) (a)n + (b a) Relating the polynomials in n on the far left and right, we have a 1 and b a 0. But then a 1/ and b a. Therefore, S (n) n i 1 n + 1 n n(). Alternatively, once one knows that S (n) is a polynomial of degree two, we could use the points (0, 0), (1, 1), and (, 3coming from the fact that S (0) 0, S (1) 1, and S () 3) and use Lagrange Interpolation to find that S (n) 0 (n 1)(n n 0)(n n 0)(n 1) n(n + 1) (1 0)( 01 0)(1 0)( 1) Proof VI: By Combinatorics Proof. Let S denote the n-element set {1,,..., n}. We count the number of ways to choose a two element subset from S. First, we can choose the first element in n ways and the second element in (n 1) ways. However, choosing i and then j produces the same two element subset as choosing j then i. So the number of ways of choosing a two element subset from S is n(n 1). Alternatively, suppose the larger of the two numbers chosen is i. Then for i, 3,..., n, there are i 1 choices for the second number j. That is for i, 3,..., n, there are 1,,, n 1 possible two element subsets of S. Then in total there are (n 1) total two element subsets of S. But then n 1 n(n 1) i (n 1) Proof VII: By Binomial Coefficients Proof. First, we prove Pascal s Identity: ( k n k + n k 1). We count the number of ways to choose a k element subset from the set {1,,..., n + 1} in two different ways. Since they count the same thing, they must be equal. First, we do this directly. The number of k element subsets one can choose from this set is exactly ( ) k. Second, each k element subset either contains n + 1 or does not. The number of k element subsets containing n + 1 is ( n k 1) while the number of k element subsets not containing n + 1 is ( n k. But then the number of k element subsets is n k + n k 1. Therefore, k n k + n k 1). We will need this identity for the starred equality below. 3 This actually takes a bit more work to show. Let V be the space of all polynomials defined over N {0} over a field F of characteristic 0. Define the forward difference operator Dp(n) : p(n + 1) p(n). If p(n) has degree d + 1, then Dp(n) has degree at most d. Let V d denote the subspace of V consisting of polynomials of degree at most d. Then we have dim F V d d + 1. Choosing the standard basis, observe that matrix for the forward difference operator is upper triangular and defines an operator D : V d+1 V d. Then the result is clear. 4

5 Now we show n i ( ) using induction. The case where n 1 is simple: 1 i 1 and ( ) 1. Assume the result is true for n 1,,..., k 1. Then k k 1 i k + i ( ) k k + 1 ( ) k + 1 k(k + 1) Then n i ( ) n() follows by induction Proof VIII: By Graph Theory Proof. A graph is a collection of vertices (one can think of these as points) and edges (one can think of these as line segments connecting the vertices, i.e. points). A graph is called complete if given any two distinct vertices in the graph, there is an edge connecting them. That is, a complete graph has the maximal number of edges. A complete graph with n vertices is denoted K n. For example, the complete graph K 6 is illustrated in Figure 1. Figure 1: The complete graph K 6. Consider the complete graph K n. Label the vertices v 1, v,..., v n. Associate to vertex v 1 the (n 1)- edges connecting it to all the other vertices in K n. Associate to vertex v the (n )-edges connecting it to all the other vertices in K n except for v 1. Continue this process for v 3, v 4,..., v n. Notice that for each i, the association for v i+1 contributes no new edges and this process never duplicates an edge. Let v i denote the number of edges associated with v i. Then the number of edges in K n is... # of edges v i v 1 + v + + v n 1 + v n (n 1) + (n )

6 But then we have n v i n 1 i. The result will follow if we can show that the number of edges, n v i, is n(n 1). But every edge in K n connects two vertices. The number of edges must then be the number of ways one can select two vertices to connect. But this is precisely ( ) n n(n 1). Therefore, we have n 1 ( ) n n(n 1) i v i But this is exactly what was to be shown. Proof IX: By Triangles Proof. (Larson [Lar85]): We want to show that (n 1) n(n 1). Represent the sum (n 1) as a triangular array of yellow circles. Place a row of n blue dots beneath this array to create larger a triangular array of dots. The case when n 5 is illustrated in Figure. Observe that if Figure : An illustration for the triangular array for n 5. one chooses any two distinct blue dots, there is a unique yellow dot in the upper portion of the triangular array associated to the pair of dots as illustrated in Figure. Vise versa for each yellow dot, there is a unique pair of blue dots associated to it. That is, there is a one-to-one correspondence between yellow dots and pairs of blue dots. But then the number of yellow dots, (n 1), must be the same as the number of ways of choosing two distinct blue dots, ( n ). Then we must have n 1 i (n 1) ( ) n n(n 1) But this was exactly what was to be shown. 6

7 Proof X: By Telescoping Series Proof. Observe that (i + 1) i (i + i + 1) i i + 1. The series (i+1) i ( 1 )+(3 )+(4 3 )+ +(() n ) 1 +() 1+(n +) n +n Take note also that n 1 n. We also have... (i + 1) i + 1 i + 1 n + so that n i n + n (i + 1). Putting these results together, we have... i n + n + i + 1 (i + 1) i n + (n + n) n + n n(n + 1) Therefore, n i n() i Proof XI: By l Hôpital Proof. Consider the finite geometric series 1 + r + r + + r n 1 r 1 r Differentiating both sides of the equality yields d ( 1 + r + r + + r n) 1 + r + 3r + + nr n 1 dr ( ) d 1 r dr 1 r (1 r)(n + 1)rn ( 1)(1 r ) (1 r) nr (n + 1)r n + 1 (1 r) We obtain the sum n by taking the limit as r tends to 1: nr (n + 1)r n + 1 L.H. n(n + 1)(r 1)r n 1 lim r 1 (1 r) lim r 1 (1 r) L.H. n(n + 1)(n(r 1) + 1)r n lim r 1 n(n + 1) 7

8 where L.H. denotes that the equality follows by application of l Hôpital s Rule. But it then follows that n i n() References [Lar85] L. Larson. A Discrete Look at n. College Mathematics Journal, 16:369 38,