Convergence of a logarithmic series

Question

Is this series $\sum\limits_{x=1}^{\infty}\frac{1}{\log(1+x)}$ convergent or divergent?

If it were only $\sum\limits_{x=1}^{\infty}\frac{1}{\log(x)}$, then I could use Cauchy condensation formula.

But now I'm stuck with it.

Answer 1

Note that you can easily find a $N$ such that for all $n \geq N$, $n > \log(1+n)$ so $\frac{1}{n} < \frac{1}{\log (1+n)}$.

Now, you know that the harmonic series diverges, so by comparison, the sum stated diverges.

Answer 2

Notice when we write

$$\sum_{x=1}^\infty \frac{1}{\log (1+x)} = \frac{1}{\log 2} + \frac{1}{\log 3} + \frac{1}{\log 4} + \cdots$$

your sum looks exactly like $\sum_{x=2}^\infty \frac{1}{\log x}$. You are allowed to reindex your sum by picking a new dummy variable, such as $y = 1+x$. To figure out your new lower bound, you simply plug in $x=1$ to get $y=2$, and similarly for your upper bound. Now you can apply the Cauchy condensation test as originally desired.