Estimate the sum $\sum_{n = 1}^ N \frac1n$ Let $N\geqslant1$ be an integer. Then which of the following statements are true ?

*

*$\displaystyle\sum\limits_{n=1}^N\dfrac1n\leqslant1+\log N$


*$\displaystyle\sum\limits_{n=1}^N\dfrac1n<1+\log N$


*$\displaystyle\sum\limits_{n=1}^N\dfrac1n\leqslant\log N$


*$\displaystyle\sum\limits_{n=1}^N\dfrac1n\geqslant\log N$
My Attempt :
$$\underbrace{\frac11+\frac11+\frac11+\frac11+\frac11+\frac11+\ldots+\frac11}_{N\text{ times}}=N$$
$$\geqslant\sum_{n=1}^N\frac1n=\frac11+\frac12+\frac13+\frac14+\frac15+\frac16+\ldots+\frac1N$$
$$\geqslant\underbrace{\frac1N+\frac1N+\frac1N+\frac1N+\frac1N+\frac1N+\ldots+\frac1N}_{N\text{ times}}$$
$$=\frac NN=1$$
$$\implies1\leqslant\sum_{n=1}^N\frac1n\leqslant N$$
Also $\;\log N\leqslant N,\;\forall N\geqslant1\,.\;$ I'm unable to combine this information to answer the question. Please help me.

$$\log(1+x)<x\;,\;\;\text{ if }x>0$$
Let $\;x=\dfrac1n\;,\;$ then
$$\frac1n>\log(1+n)-\log n$$
Next, by telescopic summing we get
$$\sum_{n=1}^N\frac1n>\log(1+N)-\log 1=\log(1+N)>\log N$$
Edit :
Let $\;f(x)=\log(1+x)-x$
$$\implies f'(x)=\frac1{1+x}-1=-\frac{x}{1+x}<0\;,\;\;\text{ if }x>0\;.$$
So $f(x)$ being a decreasing function of $\;x\;$ for $\;x\geqslant0\;,$
$$f(x)<f(0)\implies\log(1+x)<x \;,\;\;\text{ if }x>0\;.$$
