# 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.

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Hint: Use left/right hand approximations to $\int_0^N\frac{1}{x}\,dx$ –  Arturo Magidin Jun 10 '12 at 2:55
Is it $\int_{0}^{N}\frac{1}{x} dx$ or $\int_{1}^{N}\frac{1}{x} dx$? –  Kns Jun 10 '12 at 2:57
@Arturo Magidin What is the meaning of left/right hand approximation? –  Kns Jun 10 '12 at 3:01
Yes, sorry, from $1$ to $N$. "Left hand approximation" means you use the value of the function on the left end of the subinterval to estimate the area, and "right hand approximation" means you use the value at the right end of the subinterval. –  Arturo Magidin Jun 10 '12 at 3:05

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.

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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.

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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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How did u write $log ( 1+x)$ in terms of $x/(1+x)$ ? –  Theorem Jun 10 '12 at 7:40
The easiest way to to use $$-\log(1-x) = x+\frac{x^2}{2}+\frac{x^3}{3}+\cdots$$ and let $x=\dfrac{x'}{x'+1}$. However, I proved the expansion differently. –  Pedro Tamaroff Jun 10 '12 at 18:29

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.

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