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

Question

Show that the following sum is divergent $$\sum_{n=1}^{\infty}\ln\left(1+\frac1n\right)$$

I thought to do this using Taylor series using the fact that $$ \ln\left(1+\frac1n\right)=\frac1n+O\left(\frac1{n^2}\right) $$ Which then makes it clear that $$ \sum_{n=1}^{\infty}\ln\left(1+\frac1n\right)\sim \sum_{n=1}^{\infty}\frac1n\longrightarrow \infty $$ But I feel like I overcomplicated the problem and would be interested to see some other solutions. Also, would taylor series be the way you would see that this diverges if you were not told?

Answer 1

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

Answer 2

$$\sum_{n=1}^{m}\log\left(\frac{n+1}{n}\right)=\sum_{n=1}^{m}(\log(n+1)-\log n)=\log(m+1)$$ The partial sums clearly diverge. Alternatively using the cauchy condensation test the series converges iff the series$$\sum_{n=1}^{\infty}2^{n}\ln(1+1/2^{n})$$ converges. The transformed series diverges since the terms don't go to zero and so the original series diverges.

Answer 3

Applying a property of logarithms gives the equality $\displaystyle \sum_{n=1}^\infty \ln(1 + 1/n) = \ln \Bigg( \prod_{n=1}^\infty (1 + 1/n) \Bigg)$. Therefore, if $\displaystyle \sum_{n=1}^\infty \ln(1 + 1/n)$ converges, say to $c \in \mathbb{R}^+$, then $\displaystyle \prod_{n=1}^\infty (1 + 1/n)$ should converge to $e^c$.

Therein lies a contradiction: expanding this product yields a clearly divergent sum: the expansion will include a positive copy of $1/n$ for all $n \in \mathbb{N}$.

Answer 4

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! .

Answer 5

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.

Answer 6

"Sophisticated" does not mean "complicated". In my opinion, despite using more sophisticated ideas (asymptotic analysis), your proof is simpler than all of the other current answers — even the one expressing it as a telescoping series.

Incidentally, you possibly made an oversight: to complete the proof,

$$ \sum_{n=1}^{\infty} O\left(\frac{1}{n^2} \right) = O\left( \sum_{n=1}^{\infty} \frac{1}{n^2} \right) = O(1)$$

(also, a remark: for this argument to be valid, it's important that the $O$ on the left is uniform; e.g. the same 'hidden constant' works for all $n$)

Answer 7

You can also use the Abel's summation $$\sum_{n=1}^{N}\log\left(1+\frac{1}{n}\right)=\sum_{n=1}^{N}1\cdot\log\left(1+\frac{1}{n}\right)=N\log\left(1+\frac{1}{N}\right)+\int_{1}^{N}\frac{\left\lfloor t\right\rfloor }{t\left(t+1\right)}dt $$ where $\left\lfloor t\right\rfloor $ is the floor function. Since $\left\lfloor t\right\rfloor =t+O\left(1\right) $ we have $$\sum_{n=1}^{N}\log\left(1+\frac{1}{n}\right)=N\log\left(1+\frac{1}{N}\right)+\log\left(N+1\right)+O\left(1\right) $$ and taking $N\rightarrow\infty$ we can see that the series diverges.