Integrals

Stuck on the deduce part of showing the inequality contains the sum of 1/n from 2 to N

(1) You can plug any number in n, i.e. n=2,3,4... etc., and the inequality is still true.
(2) You can add inequalities. i.e. if a>b and c>d, then a+c>b+d.

I'm assuming you're stuck at the start. If you know these facts, the deduction should be fairly straightforward.
In fact, it's kind of a common "trick" involving sums and inequalities.

EDIT: Here's what you can do:

What do you think your problem is? Is it the 1-1/N on the right rather than 1?

Yes, are you going to say that N tends to infinity 1-1/N tends to 1 to get that form.
But it isn't stated that N is a big number though?

The only other alternative that I can think of is to use 1-1/N< 1 but then the inequality signs will be different if you sub it in

Yes, are you going to say that N tends to infinity 1-1/N tends to 1 to get that form.
But it isn't stated that N is a big number though?

The only other alternative that I can think of is to use 1-1/N< 1 but then the inequality signs will be different if you sub it in

I'm now confused again haha, from my above post I have ln(N) - sum <=1-1/N which isn't what they asked me to deduce. Hence I said that 1-1/N < 1 for all natural numbers of N however that'll result to my inequality to be ln(N) -sum < 1 which is what I'm confused about since I have the inequality sign <1 not <=1 which is what they want me to show.