Proof by Induction: Recursive Sum

Question

We have a sum $S_n$ defined by: $$S_1 = \frac{1}{1\cdot3}$$ $$S_2 = \frac{1}{1\cdot3}+\frac{1}{3\cdot5}$$ $$S_3 = \frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}$$ $$S_N = \frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{(2n-1)(2n+1)}$$

We have to figure out a formula for such a sum which I guessed to be $$S_N = S_{N-1}+\frac{1}{(2n-1)(2n+1)}$$

And then we have to prove the formula is correct by induction. To be honest, I don't even know if my formula is correct. Any help would be appreciated.

Answer 1

We can rewrite each term using the following:

$$\frac{1}{(2n-1)(2n+1)} = \frac12 \left(\frac{1}{2n-1} - \frac{1}{2n+1}\right)$$

Thus

\begin{eqnarray} S_n &=& \frac{1}{1\cdot 3} + \frac{1}{3\cdot 5} + \frac{1}{5\cdot 7} + \cdots + \frac{1}{(2n-1)(2n+1)}\\ &=& \frac12\left(\frac11 - \frac13\right) + \frac12\left(\frac13 - \frac15\right) + \frac12\left(\frac15 - \frac17\right) + \cdots + \frac12\left(\frac{1}{2n-1} - \frac{1}{2n+1}\right) \\ &=& \frac12\left[1 + \left(- \frac13 + \frac13 \right)+ \left(- \frac15 + \frac15\right)+\left( - \frac17 + \cdots + \frac{1}{2n-1}\right) - \frac{1}{2n+1}\right]\\ &=& \frac12\left(1 + 0 + 0 + \cdots + 0 + \frac{1}{2n+1}\right)\\ &=& \frac12 - \frac{1}{2(2n+1)} \end{eqnarray}

As $n \to \infty$, $S_n \to \frac12$.

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To show the first equality,

\begin{eqnarray} \frac12 \left(\frac{1}{2n-1} - \frac{1}{2n+1}\right) &=& \frac12 \left(\frac{(2n+1) - (2n-1)}{(2n-1)(2n+1)}\right)\\ &=& \frac12 \left(\frac{2}{(2n-1)(2n+1)}\right) = \frac{1}{(2n-1)(2n+1)} \end{eqnarray}

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Edit: Given the formula $S_n = \frac12 - \frac{1}{2(2n+1)}$, we can prove it using induction as follows:

1. Base case: $n=1$

$$S_1 = \frac12 - \frac{1}{2(2+1)} = \frac12 - \frac16 = \frac13$$

2. Assume the forumla holds up to $S_n$ and prove for $S_{n+1}$:

\begin{eqnarray} S_{n+1} &=& S_n + \frac{1}{(2n+1)(2n+3)}\\ &=& \frac12 - \frac{1}{2(2n+1)} + \frac{1}{(2n+1)(2n+3)}\\ &=& \frac12 - \frac{2n+3 - 2}{2(2n+1)(2n+3)}\\ &=& \frac12 - \frac{2n+1}{2(2n+1)(2n+3)}\\ &=& \frac12 - \frac{1}{2(2n+3)}\\ &=& \frac12 - \frac{1}{2(2(n+1)+1)}\\ \end{eqnarray}

Therefore the formula holds for all $n \in \Bbb N$ by induction.