Prove $\frac n2 > \sum_{k=1}^n (1/k) -1$

Question

This is the problem: $$ \frac n2 > \sum_{k=1}^n (1/k) -1$$

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Here's my try.

P(1): $$\frac12 > \frac11 -1$$

$$\frac12 > 0$$

P(n) => P(n+1): $$\sum_{k=1}^{n+1} \frac1k - 1 = \sum_{k=1}^n \frac1k + \frac 1{n+1} - 1 < \frac n2 + \frac 1{n+1}$$

Now the right side of inequality is bigger than sum to $n+1$ which is bigger than sum to $n$ that means that it's bigger than sum to $n$.

We have $$\frac n2+\frac1{n+1} > \sum_{k=1}^n \frac1k - 1$$

I don't know what to do next. I wrote this above to help you to help me with this kind of problems. I get confused when i need to prove inequality using induction.

Thanks!

Answer 1

You almost have a complete proof.

The key point that you need to finish the proof is that if $n \ge 1$ then $\dfrac{1}{n+1} \le \dfrac{1}{1+1} = \dfrac{1}{2}$.

Hence, your inductive step $P(n) \implies P(n+1)$ can be completed as follows: $$\sum_{k=1}^{n+1} \frac1k - 1 = \sum_{k=1}^n \frac1k + \frac 1{n+1} - 1 < \frac n2 + \frac 1{n+1} \le \dfrac{n}{2}+\dfrac{1}{2} = \dfrac{n+1}{2}.$$