Answer
$\displaystyle\frac{1}{1\cdot 2}+\displaystyle\frac{1}{2\cdot 3}+\displaystyle\frac{1}{3\cdot 4}+\ldots+\displaystyle\frac{1}{n\left(n+1\right)}=\displaystyle\frac{n}{n+1}$.
See the proof below.
Work Step by Step
Consider the sequence $\displaystyle\frac{1}{1\cdot 2},\displaystyle\frac{1}{2\cdot 3},\displaystyle\frac{1}{3\cdot 4},\displaystyle\frac{1}{4\cdot 5},\ldots$
For each natural number $n$ let
$$S_n=\displaystyle\frac{1}{1\cdot 2}+\displaystyle\frac{1}{2\cdot 3}+\displaystyle\frac{1}{3\cdot 4}+\ldots+\displaystyle\frac{1}{n\left(n+1\right)}.$$
Let's calculate a few values of $S_n$ so we can look for a pattern and then propose a formula for $S_n$.
Observe that
$$S_1=\displaystyle\frac{1}{1\cdot 2}=\displaystyle\frac{1}{2}=\displaystyle\frac{1}{1+1}$$
$$S_2=\displaystyle\frac{1}{1\cdot 2}+\displaystyle\frac{1}{2\cdot 3}=\displaystyle\frac{1}{2}+\displaystyle\frac{1}{6}=\displaystyle\frac{2}{3}=\displaystyle\frac{2}{2+1}$$
and
$$S_3=\displaystyle\frac{1}{1\cdot 2}+\displaystyle\frac{1}{2\cdot 3}+\displaystyle\frac{1}{3\cdot 4}=\displaystyle\frac{1}{2}+\displaystyle\frac{1}{6}+\displaystyle\frac{1}{12}=\displaystyle\frac{3}{4}=\displaystyle\frac{3}{3+1}.$$
The previous pattern leads us to suspect that for any positive integer it is true that
$$S_n=\displaystyle\frac{n}{n+1}.$$
We will use mathematical induction to show that our guess is correct.
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In order to show that the formula holds we must show that it is true for $n=1$ and that if it is true for $n=k$ then, it is true for $n=k+1$.
$\textbf{1.}$ The formula holds for $n=1$ since
$S_1=\displaystyle\frac{1}{1\cdot 2}=\displaystyle\frac{1}{2}=\displaystyle\frac{1}{1+1}$.
$\textbf{2.}$ Suppose the formula holds for $n=k$, that is, suppose that
$$S_k=\displaystyle\frac{1}{1\cdot 2}+\displaystyle\frac{1}{2\cdot 3}+\displaystyle\frac{1}{3\cdot 4}+\ldots+\displaystyle\frac{1}{k\left(k+1\right)}=\displaystyle\frac{k}{k+1}.$$
Then, adding $\displaystyle\frac{1}{\left(k+1\right)\left(\left(k+1\right)+1\right)}$ to both sides of the equality
$$\displaystyle\frac{1}{1\cdot 2}+\displaystyle\frac{1}{2\cdot 3}+\displaystyle\frac{1}{3\cdot 4}+\ldots+\displaystyle\frac{1}{k\left(k+1\right)}=\displaystyle\frac{k}{k+1}$$
gives us that
$$\displaystyle\frac{1}{1\cdot 2}+\displaystyle\frac{1}{2\cdot 3}+\displaystyle\frac{1}{3\cdot 4}+\ldots+\displaystyle\frac{1}{k\left(k+1\right)}+\displaystyle\frac{1}{\left(k+1\right)\left(\left(k+1\right)+1\right)}=$$
$$\displaystyle\frac{k}{k+1}+\displaystyle\frac{1}{\left(k+1\right)\left(\left(k+1\right)+1\right)}$$
Now, observe that
$$\displaystyle\frac{k}{k+1}+\displaystyle\frac{1}{\left(k+1\right)\left(\left(k+1\right)+1\right)}=\displaystyle\frac{k}{k+1}+\displaystyle\frac{1}{\left(k+1\right)\left(k+2\right)}=$$
$$\displaystyle\frac{k\left(k+2\right)}{\left(k+1\right)\left(k+2\right)}+\displaystyle\frac{1}{\left(k+1\right)\left(k+2\right)}=\displaystyle\frac{k^2+2k+1}{\left(k+1\right)\left(k+2\right)}=$$
$$\displaystyle\frac{\left(k+1\right)^2}{\left(k+1\right)\left(k+2\right)}=\displaystyle\frac{k+1}{k+2}=\displaystyle\frac{k+1}{\left(k+1\right)+1}.$$
Thus,
$$S_{k+1}=\displaystyle\frac{1}{1\cdot 2}+\displaystyle\frac{1}{2\cdot 3}+\displaystyle\frac{1}{3\cdot 4}+\ldots+\displaystyle\frac{1}{k\left(k+1\right)}+\displaystyle\frac{1}{\left(k+1\right)\left(\left(k+1\right)+1\right)}=\displaystyle\frac{k+1}{\left(k+1\right)+1}$$
and therefore, the formula holds for $n=k+1$ whenever it is true for $n=k$.
Hence, the mathematical induction principle guarantees that the formula is true for every positive integer $n$.
