Bounds for the Harmonic k-th partial sum.

Question

I need to bound the k-th partial sum or the Harmonic series. i.e.

$$ln(k+1)<\sum_{m=1}^{k}\frac{1}{m}<1+ln(k)$$

I'm triying to integrate in $[m,m+1]$ in the relation $\frac{1}{m+1}<\frac{1}{x}<\frac{1}{m}$ for all $x\in[m,m+1]$ and I get: $$\int_{m}^{m+1}\frac{1}{m+1}dx<ln(m+1)-ln(m)<\int_{m}^{m+1}\frac{1}{m}dx$$ then $$\sum_{m=1}^{k}\frac{1}{m+1}<ln(k)<\sum_{m=1}^{k}\frac{1}{m}$$ but I don know how conclude or continue... please help.

Answer 1

For $m\ge 2$, $1/u\le 1/m \le 1/(u-1)$ for $u \in [m,m+1]$

By integration on this interval

$$\ln(m+1)-\ln m \le 1/m \le \ln m - \ln(m-1)$$

You then just have to sum those inequalities.

Answer 2

An intuitive approach here can be found by thinking pictorally about the sums and functions at hand. We know that the area underneath the curve $\frac{1}{x}$ from $1$ to $k$ is given by $\ln(k)$. The partial sums $\sum_{m=1}^n \frac{1}{m}$ can be visualized by laying out a block of height and length $1$ starting at $x = 1$, then of height $\frac{1}{2}$ and length $1$ starting at $x = 2$, of height $\frac{1}{3}$ and length $1$ at $x = 3$, and so on up to a block of height $\frac{1}{k}$ and length $1$ at $x = k$. If we draw out a few blocks and then think about how to bound above or below by graphs that look like possibly shifted versions of $\frac{1}{x}$, why this bound holds should become much easier to see. A rigorous argument with integrals then follows fairly naturally.

Answer 3

Noting that $$ \frac1{x+1}<\frac1m<\frac{1}{x},x\in(m-1,m)$$ one has $$ \sum_{m=1}^k \frac1m=\sum_{m=1}^k\int_{m-1}^m\frac{1}{m}dx<1+\sum_{m=2}^k\int_{m-1}^m\frac{1}{x}dx=1+\int_1^k\frac1xdx=1+\ln k$$ and $$ \sum_{m=1}^k \frac1m=\sum_{m=1}^k\int_{m-1}^m\frac{1}{m}dx>\sum_{m=1}^k\int_{m-1}^m\frac{1}{x+1}dx=\int_0^1\frac1{x+1}dx+\int_1^2\frac1{x+1}dx+\dots+\int_{k-1}^k\frac1{x+1}dx=\int_0^k\frac1{x+1}dx=\ln (k+1). $$ So $$ \ln(k+1)<\sum_{m=1}^k \frac1m<1+\ln k. $$

Answer 4

Continuing your original thinking with a few modifications:

$$ln(m+1)-ln(m)<\int_{m}^{m+1}\frac{1}{m}dx$$ then $$\sum_{m=1}^{k}ln(m+1)-ln(m)<\sum_{m=1}^{k} \int_m^{m+1}\frac{1}{m} dx$$ and $$ln(k+1)<\sum_{m=1}^{k}\frac{1}{m}$$

which gives you the LHS of the inequality.

The RHS comes from:

$$\frac{1}{m-1} < x < \frac{1}{m}$$ for $x\in [\frac{1}{m}, \frac{1}{m - 1}]$

Therefore:

$$\int_{m-1}^m \frac{1}{m} dx< ln(m) - ln(m-1)$$

and

$$\sum_{m=2}^{k}\int_{m-1}^m \frac{1}{m} dx< \sum_{m=2}^{k} ln(m) - ln(m-1)$$

then

$$\sum_{m=2}^{k}\frac{1}{m} < ln(k)$$

and finally,

$$\sum_{m=1}^{k}\frac{1}{m} < ln(k) + 1$$

Where we added $1$ to each side.

Answer 5

In order to arrive at this inequality, we can use an approach that uses your ideas. We have for every $m$,

$$\dfrac{1}{m+1} < \ln(m+1) - \ln(m) < \dfrac{1}{m}$$

Now consider the right hand side:

$$ \ln(m+1) - \ln(m) < \dfrac{1}{m}$$

Putting $m=k$, we can arrive at the left hand side inequality as follows:

$$ \begin{align}\ln(k+1) &< \dfrac{1}{k} + \ln(k) \\ &< \dfrac{1}{k}+ \dfrac{1}{k-1} + \ln(k-1) \tag{putting $m=k-1$} \\ &\vdots \\ &< \sum^{k}_{i=1} \dfrac{1}{i} \end{align}$$

This gives the lower bound of the required inequality.

Now consider,

$$\dfrac{1}{m+1} < \ln(m+1) - \ln(m)$$ $$ \implies \dfrac{1}{m+1} + \ln(m) < \ln(m+1)$$

Here we will start with $m=k-1$:

$$\begin{align} \ln(k) &> \dfrac{1}{k} + \ln(k-1) \\ &> \dfrac{1}{k}+ \dfrac{1}{k-1}+\ln(k-2) \tag{putting $m=k-2$} \\ &\vdots \\ &> \sum^{k}_{i=2} \dfrac{1}{i} + \ln(1) \\ &= \sum^{k}_{i=2} \dfrac{1}{i} \end{align}$$

Thus, adding 1 on both sides gives us the desired upper bound of the inequality.

QED.