Ways of showing $\sum_\limits{n=1}^{\infty}\ln(1+1/n)$ to be divergent 
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?
 A: 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.
A: 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!
.
A: 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 $n \in \mathbb{N}$.
A: $$\sum_{n=1}^{m}\log\left(\frac{n+1}{n}\right)=\sum_{n=1}^{m}(\log(n+1)-\log n)=\log(m+1)$$
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.
A: Notice the following:
$$\log\left(1+\frac{1}{n}\right)=\log\left(\frac{n+1}{n}\right)=\log(n+1)-\log(n)$$
Hence $$\sum_{k=1}^{n}\log\left(1+\frac{1}{k}\right)=\log(n+1) \to \infty$$
A: In THIS ANSWER, I showed using elementary analysis only that the logarithm function satisfies the inequalities 
$$\frac{x-1}{x}\le\log(x)\le x-1\tag1$$
Letting $x=1+\frac1n$ in $(1)$ reveals 
$$\frac1{n+1}\le\log\left(1+\frac1n\right)\le \frac1n\tag2$$
And using the left-hand side of $(2)$ shows that
$$\sum_{n=1}^N \log\left(1+\frac1n\right)\ge \sum_{n=1}^N \frac1{n+1}$$
Inasmuch as the harmonic series diverges, we see that the series of interest diverges also.  And we are done!
A: "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 $n$)
A: 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.
