Proof by induction using summation 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.
Thanks in advance!
 A: 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.
A: 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
$$
\frac{n+1}{n+2}
$$
A: 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 $n \geq 1$
