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

Question

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 .

Answer 1

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.

Answer 2

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

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

Answer 3

$$\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)=\prod_{k=1}^{n-1}\left(\frac{k+1}{k}\right)=n $$ 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 $H_n>\log(n)$.

Answer 4

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

Answer 5

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 $$ \sum_{k=1}^{n}\frac{1}{k} \geq \int_{x = 1}^{n+1}\frac{1}{x} = \log (n+1) > \log (n). $$