Show that $\sum_{n=1}^{\infty} \dfrac{1}{n} \ln \left( 1+\dfrac{x}{n} \right)$, with $x>-1$ is pointwise convergent. Title says it all, really. I have to show, that:
 $\sum_{n=1}^{\infty} \dfrac{1}{n} \ln \left( 1+\dfrac{x}{n} \right)$, with $x&gt;-1$
is pointwise convergent, but I have no idea where to start. I am thinking, that I have to show it by comparison, but I am lost on what to compare it with.

Anyone got any ideas?

Much appreciated.
Hint: When $x&gt;0$ $$0&lt;\ln\left(1 + \frac{x}{n}\right) &lt; \frac{x}{n}$$ for all positive $n$, and when $-1\le x&lt;0$ $$0&gt;\ln\left(1 + \frac{x}{n}\right) &gt; \frac{x}{n-1} $$ for $n&gt;1$.
$1+x \leq e^x$ $\forall x \in R$ so $(1+x/n)^{1/n} \leq e^{x/n^2}$ hence $1/n \ln (1 + x/n) \leq x/n^2$.