Proof by Induction for $\sum_{i = 1}^{n - 1} \frac{1}{i(i+1)}$

Question

$$\sum_{i = 1}^{n - 1} \frac{1}{i(i+1)} = \frac{n- 1}{n}$$

How would you go about proving this by induction? the base case $n=1$ gives you $1/2 = 0$

The only thing I can think of using is $\sqrt{2}$, but that is not a natural number.

Answer 1

Your understanding of the base case is incorrect. When $n = 1$, the sum on the left-hand side is empty, because the upper index is $n - 1 = 0$, meaning there are no terms to add. This gives $0 = 0$ as desired.

The induction step is straightforward. If we let $$S_n = \sum_{i=1}^{n-1} \frac{1}{i(i+1)},$$ then $$S_{n+1} = S_n + \frac{1}{n(n+1)}.$$ So if the induction hypothesis $$S_n = \frac{n - 1}{n}$$ is true, what does that say about $S_{n+1}$?

Answer 2

When $n = 1$, the sum on the left side is empty, because the upper index is $n - 1 = 0$, that is there are no terms to add. This gives $0 = 0$ as desired. Suppose that $$S_n = \sum_{i=1}^{n-1} \frac{1}{i(i+1)},$$ then we have to prove $$S_{n+1} = \frac{n}{ (n+1)}.$$ So if the induction hypothesis $$S_n = \frac{n - 1}{n}$$ is true, we have: $$S_{n+1} = S_n+ \frac{1}{n (n+1)}.$$ that is $$S_{n+1} =\frac{n-1}{n}+ \frac{1}{n (n+1)}=\frac{n}{n+1}.$$ The proof is compelete.

Answer 3

The $2$ answers above are indeed excellent, and should be sufficient for your problem. Below I offer a different way of proving this result: $$\begin{align}\sum_{i=1}^{n-1}\frac{1}{i(i+1)}&=\sum_{i=1}^{n-1}\frac{1}{i}-\frac{1}{i+1}\\ &=\frac{1}{1}-\frac{1}{2}\\ &+\frac{1}{2}-\frac{1}{3}\\ &+\frac{1}{3}-\frac{1}{4}\\ &~~\vdots\\ &+\frac{1}{n-1}-\frac{1}{n}\end{align}$$ You should be able to see that all terms cancel out other than $1$ and $-\frac{1}{n}$ Hence, $$\sum_{i=1}^{n-1}\frac{1}{i(i+1)}=1-\frac{1}{n}=\frac{n-1}{n}$$ as required.