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I am attempting to prove by induction that $\frac{1}{1\times2} + \frac{1}{2\times3} + ... + \frac{1}{n(n+1)} < 1$ for all integers $n \geq 1$.

I have been able to show (using a telescoping sum) that the LHS is equal to $1 - \frac{1}{n+1}$, which is clearly less than 1 for all positive integers $n$.

But I am trying to use induction to prove the above inequality without using the explicit formula. What I have done so far:

$\frac{1}{1\times2} + \frac{1}{2\times3} + ... + \frac{1}{n(n+1)} + \frac{1}{(n+1)(n+2)} < 1 + \frac{1}{(n+1)(n+2)}$

From here I have no idea how to prove the inequality.

Is it possible to prove this inequality using induction or is finding an explicit formula for the LHS the only way?

1123581321
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    The problem is that it is just not true that if a quantity is less than $1$ then that quantity plus $\frac{1}{n(n+1)}$ remains less than $1$. You can't prove this statement by induction without strengthening it, and you can strengthen it all the way up to an exact formula, which works fine. – Qiaochu Yuan Dec 30 '20 at 07:36
  • Prove by induction that the $n$th term is $1-\frac{1}{n+1}$. – Gary Dec 30 '20 at 07:38
  • Do you mean the $n$th term of the series or the $n$th partial sum of the series? – 1123581321 Dec 30 '20 at 07:40
  • I ment the partial sum. – Gary Dec 30 '20 at 07:44
  • Use ${1\over x(x+1)}={1\over x}-{1\over x+1}$, so your sum will be simple to compute.. – Thomas Dec 30 '20 at 09:39

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