Prove that $\sum_{i=j}^k \frac 1i$ is $O(\ln(k)-\ln(j-1))$ using induction for all $i$.
The way I understand this question, it's nonsense - $i$ is the iteration variable, not something that can be inducted on. I think it may be possible to do by induction on $j$ and $k$, but that's not "for all $i$".
On
And to complete what Matt Samuel wrote,
Since $\frac1{x}$ is a decreasing function, and since $(\ln(x))' = \frac1{x} $, if $i \ge 2$, $\frac1{i} <\int_{i-1}^{i} \frac{dx}{x} =\ln(i)-\ln(i-1) < \frac1{i-1} $ .
Summing from $2$ to $k$, since $\sum_{i=2}^k (\ln(i)-\ln(i-1)) =\ln(k) $, $\sum_{i=2}^k \frac1{i} < \ln(k) < \sum_{i=2}^k \frac1{i-1} $.
From the left side $\sum_{i=1}^k \frac1{i} < \ln(k)+1 $.
From the right side, $\sum_{i=1}^k \frac1{i} > \ln(k)+\frac1{k} > \ln(k) $.
These should be enough to get what you want.
As is often the case for me, nothing here is original.
Hint: It is sufficient to know that $$\sum_{i=1}^n{\frac{1}{i}}=\Theta(\ln n)$$