Series

question maths short

Question

  1. Prove that the sum $ 1 + 2 + 3 + ... + n = \frac{1}{2} n(n+1) $.
  2. Prove that the sum $ 1^2 + 2^2 + 3^2 + ... + n^2 = \frac{1}{6} n(n+1)(2n+1) $.
  3. What is the value of $ 1 - x + x^2 - x^3 + ... $?

Hints

The first two equations can be proved using induction, but there is an alternate method to proving the first one.
You can solve (3) with standard results of a geometric progression, but it can also be solved in a similar fashion to (1).


Answer

Both of these series can be proven with induction. However, the first can also be proven as follows: $$ S = 1 + 2 + 3 + ... + (n-2) + (n-1) + n $$ $$ S = n + (n-1) + (n-2) + ... + 3 + 2 + 1 $$ $$ S + S = (n+1) + (n-1+2) + (n-2+3) + ... + (n-2+3) + (n-1+2) + (n+1) $$ $$ 2S = n(n+1) $$ $$ S = \frac{1}{2} n(n+1) $$

For the final sum: $$ S = 1 - x + x^2 - x^3 + ... $$ $$ xS = x - x^2 + x^3 - x^4 + ... $$ $$ S + xS = 1 + (x-x) + (x^2-x^2) + (x^3-x^3) + ... $$ $$ S + xS = 1 $$ $$ S = \frac{1}{1+x} $$ Note that this is only valid for $|x| < 1 $, as we assumed that $x^\infty$ is infinitessimally small. If $|x| > 1 $, then sum is divergent (as it alternates between $\pm \infty $)