<s>[INST] Showing inequality for harmonic series. I want to show that $$\log N<\sum_{n=1}^{N}\frac{1}{n}<1+\log N.$$ But I don't know how to show this.
 [/INST] It is a good excercise to show that
$$\tag 1 \frac{x}{{x + 1}} \leqslant \log \left( {1 + x} \right) \leqslant x$$
One alternative is expanding $\log$ into powers of $x$ and $\dfrac{x}{x+1}$, which gives
$$\eqalign{
  & \log \left( {1 + x} \right) = x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} - \frac{{{x^4}}}{4} +-  \cdots   \cr 
  & \log \left( {1 + x} \right) = \frac{x}{{x + 1}} + \frac{1}{2}{\left( {\frac{x}{{x + 1}}} \right)^2} + \frac{1}{3}{\left( {\frac{x}{{x + 1}}} \right)^3} + \frac{1}{4}{\left( {\frac{x}{{x + 1}}} \right)^4} +  \cdots  \cr} $$
From there we immediately get
$$\frac{x}{{x + 1}} \leqslant \log \left( {1 + x} \right) \leqslant x$$
Now let $x=\dfrac 1 N$
$$\eqalign{
  & \frac{{\frac{1}{N}}}{{\frac{1}{N} + 1}}\leq\log \left( {1 + \frac{1}{N}} \right)\leq\frac{1}{N}  \cr 
  & \frac{1}{{N + 1}}\leq\log \left( {\frac{{N + 1}}{N}} \right)\leq\frac{1}{N}  \cr 
  & \frac{1}{{N + 1}}\leq\log \left( {N + 1} \right) - \log N\leq\frac{1}{N} \cr} $$
Now sum from $N=1$ to $N=M$
$$\sum\limits_{N = 1}^M {\frac{1}{{N + 1}}} \leq\log \left( {M + 1} \right)\leq\sum\limits_{N = 1}^M {\frac{1}{N}} $$
$$\sum\limits_{N = 1}^{M + 1} {\frac{1}{N}}  - 1 \leqslant \log \left( {M + 1} \right) \leqslant \sum\limits_{N = 1}^M {\frac{1}{N}} $$
This gives
$$\sum\limits_{N = 1}^{M + 1} {\frac{1}{N}}  \leqslant \log \left( {M + 1} \right) + 1$$
$$\log \left( {M + 1} \right) \leqslant \sum\limits_{N = 1}^M {\frac{1}{N}}  \leqslant \sum\limits_{N = 1}^{M + 1} {\frac{1}{N}} $$
Another way to prove $(1)$ is to start from the alternative definition:
$$\log x = \mathop {\lim }\limits_{k \to +\infty} k({{x^{1/k}} - 1})$$
The note that, for $0<y<1$ and $y>1$ respectively
$$\eqalign{
  & \sum\limits_{v = 0}^{k - 1} {{y^v}}  \leqslant \sum\limits_{v = 0}^{k - 1} {1 = k}   \cr 
  & \sum\limits_{v = 0}^{k - 1} {{y^v}}  \geqslant \sum\limits_{v = 0}^{k - 1} {1 = k}  \cr} $$
Thus for $y > 0$
$${y^k} - 1 = \left( {y - 1} \right)\sum\limits_{v = 0}^{k - 1} {{y^v}}  \geqslant k\left( {y - 1} \right)$$
We now let $y^k =x$ and we get
$$x - 1 = \left( {y - 1} \right)\sum\limits_{v = 0}^{k - 1} {{y^v}}  \geqslant k\left( {{x^{1/k}} - 1} \right)$$
From this is it trivial to get 
$$\log x \geqslant 1 - \frac{1}{x}$$
since
$$\log x =  - \log \frac{1}{x} \geqslant  - \left( {\frac{1}{x} - 1} \right) = 1 - \frac{1}{x}$$
This last exposition is due to Edmund Landau.
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