proof that $ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n(n-1)} = \frac{3}{2} - \frac{1}{n} $ 
 Proof that 
$$ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n(n-1)} = \frac{3}{2} - \frac{1}{n} $$
by induction. 


 Proof 
Base case: Statement clearly holds for $n = 1$. Now assume that statement holds for some $n = k$ and lets show that it implies $n = k + 1$ holds. The proof:
$$ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n(n-1)} + \frac{1}{n(n+1)} = \frac{3}{2} - \frac{1}{n} + \frac{1}{n(n+1)} = \frac{3}{2} - \frac{1}{n} + \frac{1}{n} - \frac{1}{n + 1} $$
$$ = \frac{3}{2} - \frac{1}{n+1} $$

Now the problem is I can't find the error. The statement doesn't clearly work for $ n = 2 $. However, the assumption seems to be correct since if I assume it's true for some $n = k$ and it is true for $ n = 1$? It shouldn't be possible to show that $p(n) \implies p(n+1)$ when $p(n)$ is true and $p(n+1)$ is false. This means that $p(n)$ has to be false in this case since when $p(n)$ is false then $p(n) \implies p(n+1)$ is tautology. The problem is I don't really see how? Isn't the whole point of induction to show that $p(n)$ is true for some specific $n = k$ (not all $n$) and then show $p(n+1)$ by assuming $p(n)$. Now when $p(n)$ is false you can show anything since it's tautology but how can you be sure $p(n)$ is true if you don't show it for all $n$? And wouldn't that defeat the purpose of induction (if you have already shown it's true for all $n$)?.
 A: The base of induction cannot be $n=1$ because then $1/(n(n-1))$ is not defined. For this sum you do not need induction. The sum is equal to $$(1-1/2)+(1/2-1/3)...(1/(n-1)-1/n)=1-1/n.$$
A: For $n=1$ the last term on the lefthand side is $\frac1{1\cdot0}$, which is undefined. The induction has to start at $n=2$, and as you say, the statement is false for $n=2$. The fact that the induction step works (after you correct the sign error in your answer, which I suspect is a typo) means that the formula $\frac32-\frac1n$ is going to give the wrong answer for every $n\ge 2$.
In fact the lefthand side is a telescoping sum,
$$\left(\frac11-\frac12\right)+\left(\frac12-\frac13\right)+\ldots+\left(\frac1{n-1}-\frac1n\right)\,,$$
and the correct righthand side is $1-\frac1n$. The induction step works because the righthand side is offset from the correct value by a constant amount, $\frac12$, for every $n$.
A: As $\frac 1{n(n-1)}$ is not defined for $n =1$ and also because the first term is $\frac 1{2\cdot 1} = \frac 1{2(2-1)}$ and so the first term is for $n = 2 > 1$, then it clearly does NOT work for $n= 1$.
If the statement were true for any $n$ it would be true for any subsequent natural numbers but it is not true for any $n$.
that statement is $\sum_{k=2}^n \frac 1{k(k-1)} = \frac 32 -\frac 1n$ and that just isn't true.
But $\sum_{k=2}^n \frac 1{k(k-1)} = 1 -\frac 1n$ is.
Note the first case is for $n = 2$ and not $n =1$.
Proof:
For $n=2$ then $\frac 1{2} = 1-\frac 12$.
And if $\sum_{k=2}^n \frac 1{k(k-1)}=  1-\frac 1n$ then
$\sum_{k=2}^{n+1} \frac 1{k(k-1)} = 1-\frac 1n + \frac 1{n(n+1)} = 1-\frac {(n+1) - 1}{n(n+1)} =1-\frac n{n(n+1)} = 1-\frac 1{n+1}$.
