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.
 A: 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.
A: Look at this picture, ignoring everything to the right of $x=6$. The shaded area is $\sum\limits_{n=1}^5\dfrac1n$, and it’s clearly larger than the area under the curve $y=\dfrac1x$ between $1$ and $5$, which is $\int_1^5\frac{dx}x\,dx=\ln 5$.
Now shift the shaded region one unit to the left. The first shaded rectangle has an area of $1$, and the remaining four shaded rectangles fit under the curve $y=\dfrac1x$ between $1$ and $5$. The total area of those four rectangles is therefore less than $\int_1^5\frac{dx}x\,dx=\ln 5$, so the total area of all five rectangles is less than $1+\ln 5$.
Now generalize.
A: A simpler way  is to use one of the consequences of the Lagrange's theorem applied on $\ln(x)$ function, namely:
$$\frac{1}{k+1} < \ln(k+1)-\ln(k)<\frac{1}{k} \space , \space k\in\mathbb{N} ,\space k>0$$
Then take $k=1,2,...,n$ values to the inequality and add them up. 
The proof is complete.
A: I think I wrote this up somewhere on this website but anyways here she is again.

From the figure, you can see that the area under the blue-curve is bounded below by the area under the red-curve from $1$ to $\infty$.
The blue-curve takes the value $\frac1{k}$ over an interval $[k,k+1)$
The red-curve is given by $f(x) = \frac1{x}$ where $x \in [1,\infty)$
The green-curve takes the value $\frac1{k+1}$ over an interval $[k,k+1)$
The area under the blue-curve represents the sum $\displaystyle \sum_{k=1}^{n} \frac1{k}$ while the area under the red-curve is given by the integral $\displaystyle \int_{1}^{n+1} \frac{dx}{x}$ while the area under the blue-curve represents the sum $\displaystyle \sum_{k=1}^{n} \frac1{k+1}$
Hence, we get $\displaystyle \sum_{k=1}^{n} \frac1{k} > \displaystyle \int_{1}^{n+1} \frac{dx}{x} = \log(n+1)$
$\log(n+1)$ diverges as $n \rightarrow \infty$ and hence
$$\lim_{n \rightarrow \infty} \displaystyle \sum_{k=1}^{n} \frac1{k} = + \infty$$
By a similar argument, by comparing the areas under the red curve and the green curve, we get $$\displaystyle \sum_{k=1}^{n} \frac1{k+1} < \displaystyle \int_{1}^{n+1} \frac{dx}{x} = \log(n+1)$$ and hence we can bound $\displaystyle \sum_{k=1}^{n} \frac1{k}$ from above by $1 + \log(n+1)$
Hence, $\forall n$, we have $$\log(n+1) < \displaystyle \sum_{k=1}^{n} \frac1{k} < 1 + \log(n+1)$$
Hence, we get $0 < \displaystyle \sum_{k=1}^{n} \frac1{k} - \log(n+1) < 1$, $\forall n$
Hence, if $a_n = \displaystyle \sum_{k=1}^{n} \frac1{k} - \log(n+1)$ we have that $a_n$ is a monotonically increasing sequence and is bounded.
Hence, $\displaystyle \lim_{n \rightarrow \infty} a_n$ exists. This limit is denoted by $\gamma$ and is called the Euler-Mascheroni constant.
It is not had to show that $\gamma \in (0.5,0.6)$ by looking at the difference in the area of these graphs and summing up the area of these approximate triangles.
