# Proof by induction using summation [duplicate]

I'm trying to figure out how to solve this equation by induction and I really don't know where to begin. I have seen some YouTube tutorials, but can't understand how I can go from $$k(k+1)$$ to $$n+1$$ in the equation. The task is:

Use induction to show that:

$$\sum_{k=1}^{n} {1 \over k(k+1)} = {n \over n+1}$$

Can someone help me solve this equation? Or give me some tips for where to start? I would really appreciate it.

• Induction aside, familiar with telescoping? Dec 14, 2019 at 22:05

To prove this you would first check the base case $$n = 1$$. This is just a fairly straightforward calculation to do by hand.

Then, you assume the formula works for $$n$$. This is your "inductive hypothesis". So we have $$\begin{equation*} \sum_{k = 1}^n \frac 1{k(k + 1)} = \frac n{n + 1}. \end{equation*}$$ Now we can add $$\frac 1{(n + 1)(n + 2)}$$ to both sides: \begin{align*} \sum_{k = 1}^{n + 1} \frac 1{k(k + 1)} &= \frac n{n + 1} + \frac 1{(n + 1)(n + 2)} \\ &= \frac{n(n + 2) + 1}{(n + 1)(n + 2)} \\ &= \frac{(n + 1)^2}{(n + 1)(n + 2)} \\ &= \frac{(n + 1)}{(n + 1) + 1} \end{align*} But this is exactly the same formula again, just with $$n$$ replaced by $$n + 1$$. So if the formula works for $$n$$, it also works for $$n + 1$$. And then, by induction, we are done.

The hint is $$\sum_{k=1}^{n+1}\frac{1}{k(k+1)} =\biggl(\,\sum_{k=1}^n\frac{1}{k(k+1)}\biggr)+\frac{1}{(n+1)(n+2)}$$ which should equal $$\frac{n+1}{n+2}$$

Define the hypothesis $$H_n$$ as $$H_n : \sum_{k = 1}^n \frac{1}{k(k+1)} = \frac{n}{n+1}$$

For $$n = 1$$, we have LHS = $$\sum \limits_{k = 1}^1 \frac{1}{k(k+1)} = \frac{1}{2}$$ and RHS = $$\frac{1}{1 + 1} = \frac{1}{2}$$ so $$H_1$$ is true.

Now suppose $$H_m$$ is true for some integer $$m \geq 1$$, that is, we have $$\sum \limits_{k = 1}^m\frac{1}{k(k+1)} = \frac{m}{m+1}$$ as a given.

Then, consider $$\sum_{k = 1}^{m+1} \frac{1}{k(k+1)}$$

$$= \frac{1}{ (m+1)(m + 2)} + \sum_{k=1}^m \frac{1}{k(k+1)}$$

By induction hypothesis, we have:

$$= \frac{1}{ (m+1)(m + 2)} + \frac{m}{m+1}$$

$$= \frac{1 + m(m+2)}{(m+1)(m+2)}$$

$$= \frac{(m+1)^2}{(m+1)(m+2)}$$

$$= \frac{m+1}{(m + 1) + 1}$$

Therefore, $$\sum_{k=1}^{m+1} \frac{1}{k(k+1)} = \frac{m+1}{(m + 1) + 1}$$

So $$H_m \implies H_{m+1}$$

Since we had $$H_1$$ true, by induction, $$H_n$$ is true for all integers $$n \geq 1$$