Ways of showing $\sum_\limits{n=1}^{\infty}\ln(1+1/n)$ to be divergent

Solution 1:

Notice the following: $$\log\left(1+\frac{1}{n}\right)=\log\left(\frac{n+1}{n}\right)=\log(n+1)-\log(n)$$ Hence $$\sum_{k=1}^{n}\log\left(1+\frac{1}{k}\right)=\log(n+1) \to \infty$$

Solution 2:

This is a special case of: Suppose $f(1)= 0, f'(1) > 0.$ Then $\sum f(1+1/n) = \infty.$

Proof: From the definition of the derivative (no Taylor necessary), we have

$$\frac{f(1+h)-f(1)}{h}= \frac{f(1+h)}{h} > \frac{f'(1)}{2}$$

for small $h>0.$ Thus

$$f(1+1/n) > \frac{f'(1)}{2}\cdot\frac{1}{n}$$ for large $n.$ By the comparison test we're done.

Solution 3:

Note that we have

$$\begin{align} \log\left(1+\frac1n\right)&=\int_n^{n+1}\frac{1}{t}\,dt\\\\ &\ge\frac1{n+1} \end{align}$$

and the harmonic series diverges.

But, suppose one forgoes that comparison and instead writes

$$\begin{align} \sum_{n=1}^{2^N-1}\int_n^{n+1} \frac{1}{t}\,dt&=\int_1^{2^N}\frac{1}{t}\,dt\\\\ &=\int_1^2 \frac{1}{t}\,dt+\int_{2}^{4}\frac{1}{t}\,dt+\dots+\int_{2^{N-1}}^{2^N}\frac{1}{t}\,dt\\\\ &\ge \frac12 (2-1)+\frac14 (4-2)+\dots +\frac{1}{2^N}(2^N-2^{N-1})\\\\ &=\frac{N}{2} \end{align}$$

which goes to $\infty$ as $N\to \infty$. And we are done! .