# To prove that $\sum_{k=1}^{n}{{1}\over {k}} \ge \log n$

To show that the series $\sum_{k=1}^{\infty} {{1}\over {k}}$ diverges ; I have to prove that $${\sum_{k=1}^{n}}{{1}\over{k}} \ge {\log n}$$ The given hint is that $${{1}\over {k}}\ge \int_k^{k+1} {1\over x} dx$$

Now,evaluating the RHS, say $$L=\int_k^{k+1} {1\over x} dx\\=\log(k+1)-\log\ k\\=\log(1+{1\over k})\\={1\over k}-{1\over{2k^2}}+{1\over {3k^3}}-......$$ How can I tell this is $L \le {1\over k}?$

If this part is proved then I guess the following is like :

$$\sum_{k=1}^n {1\over k}\ge\int_1^2{1\over x}dx +\int_2^3{1\over x}dx +....\int_n^{n+1} {1\over x} dx = \int_1^{n+1} {1\over x} dx=log(n+1)$$

How to reach the conclusion $?$ . This is going nowhere .

• Most easily, use that $\frac{1}{x} \leqslant \frac{1}{k}$ for $x \geqslant k$. – Daniel Fischer Oct 17 '15 at 13:50
• @DanielFischer : Where $?$ – user118494 Oct 17 '15 at 13:56
• $$\int_k^{k+1} \frac{1}{x}\,dx \leqslant \int_k^{k+1} \frac{1}{k}\,dx = \frac{1}{k}$$ – Daniel Fischer Oct 17 '15 at 13:58
• – Martin Sleziak Jun 20 '17 at 17:53

\begin{align}\frac 1k&\geq\int_k^{k+1} \frac 1xdx\\ \text{Sum from k=1 to n:}&\\ \sum_{k=1}^n\frac 1k&\geq \sum_{k=1}^n\int_k^{k+1}\frac 1xdx=\int_1^{n+1}\frac 1xdx=\log(n+1)\geq\log n \\ \sum_{k=1}^n\frac 1k&\geq\log n\quad\blacksquare \end{align}

• ok. So when I take $lim\ \ n\rightarrow \infty$ , on the right I get $log \infty$ which is strictly positive. – user118494 Oct 17 '15 at 15:47
• Yes, $\log n \to \infty$ as $n\to \infty$. So with this you can conclude that the harmonic series diverges. – hypergeometric Oct 17 '15 at 15:57

Hint: Use the telescopic nature of the expression $\log(k+1)-\log(k)$.

For example when $n=3$, you have shown that

$\frac{1}{1}+\frac{1}{2}+\frac{1}{3} \geq \log(2)-\log(1)+\log(3)-\log(2)+\log(4)-\log(3) = \log(4)-\log(1)$.

Take the limit as $n \to \infty$ and observe the expression on the right hand side of the inequality.

• I can't see anything – user118494 Oct 17 '15 at 14:48

Hint $k \leq x \leq k+1$ thus $\frac{1}{x} \leq \frac{1}{k}$.

$$L=\int_k^{k+1} {1\over x} dx \leq \int_k^{k+1} {1\over k} dx$$

$$\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)=\prod_{k=1}^{n-1}\left(\frac{k+1}{k}\right)=n$$ hence by taking logs and exploiting $\log(1+x)\leq x$ for $|x|<1$ we have: $$\log(n) = \sum_{k=1}^{n-1}\log\left(1+\frac{1}{k}\right) \leq \sum_{k=1}^{n-1}\frac{1}{k}$$ so $H_n>\log(n)$.

The map $x \mapsto 1/x$ is stricly decreasing and the map $\log$ is strictly increasing , implying that for every $n \geq 1$ we have $$\sum_{k=1}^{n}\frac{1}{k} \geq \int_{x = 1}^{n+1}\frac{1}{x} = \log (n+1) > \log (n).$$