Question: I am learning about exact sequences that split, in the context of modules.
In this context, as I understand it, sequences that split on the left are the same as sequences that split on the right.
But in non Abelian groups, is there an easy example of an exact sequence
$1 \rightarrow G \rightarrow H \rightarrow K \rightarrow 1$ that splits on the right and not on the left, and one that splits on the left but not on the right?

Answer: 
Since $t\to\sqrt{t+1}$ is concave, it follows that, for $t\in [0,1]$,
$$1+(\sqrt{2}-1)t\leq \sqrt{t+1}\leq \sqrt{2}.$$
Hence 
$$\frac{1}{n+1}+\frac{\sqrt{2}-1}{n+2}\leq \int\_{0}^{1} t^n \sqrt{t+1} dt\leq \frac{\sqrt{2}}{n+1}.$$
It remains to show that
$$
\frac{\sqrt{2}}{n+1} - \frac{1}{2n^2}\leq
\frac{1}{n+1}+\frac{\sqrt{2}-1}{n+2}$$
that is
$$\frac{\sqrt{2}-1}{(n+1)(n+2)}\leq \frac{1/2}{n^2}$$
which holds for all $n\geq 1$.
$\sqrt{t+1}$ is approximately constant on $(0,1)$, hence the following application of Cauchy-Schwarz

$$ I\_n=\int\_{0}^{1}t^n\sqrt{1+t}\,dt\leq \sqrt{\int\_{0}^{1}t^{n}\,dt\int\_{0}^{1}t^n(1+t)\,dt}=\color{red}{\frac{1}{n+1}\sqrt{\frac{2n+3}{n+2}}}<\frac{\sqrt{2}}{n+1}$$
is expected to provide a tight upper bound. Indeed,
$$ \frac{\sqrt{2}}{n+1}-I\_n = \int\_{0}^{1}t^n\left(\sqrt{2}-\sqrt{1+t}\right)\,dt = \int\_{0}^{1}(1-t)^n\left(\sqrt{2}-\sqrt{2-t}\right)\,dt $$
leads to
$$\begin{eqnarray\*}\frac{\sqrt{2}}{n+1}-I\_n = \int\_{0}^{1}\frac{t(1-t)^n}{\sqrt{2}+\sqrt{2-t}}\,dt&\leq&\frac{1}{\sqrt{2}+1}\int\_{0}^{1}t(1-t)^n\,dt\\&=&\frac{1}{\sqrt{2}+1}\cdot\frac{1}{(n+1)(n+2)}\\&<&\frac{1}{2(n+1)^2}.\end{eqnarray\*}$$
Actually we have
$$ I\_n = \frac{\sqrt{2}}{n+1}+\frac{C\_n}{(n+1)^2} $$
where $C\_n\in\left[\frac{1}{2\sqrt{2}},\frac{1}{\sqrt{2}+1}\right]$.