<s>[INST] 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 $n$)?.
 [/INST] 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 $$(1-1/2)+(1/2-1/3)...(1/(n-1)-1/n)=1-1/n.$$
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