Three series with computed sums

(1) For $n\ge1$,
$$\frac{u_{n+1}}{u_n}=\frac{2^{n+2}}{3^{n+1}}\cdot\frac{3^n}{2^{n+1}}=\frac{2}{3}<1,$$
so by d'Alembert's ratio test the series converges. Moreover,
$$\sum_{n\ge1}\frac{2^{n+1}}{3^n}=2\sum_{n\ge1}\left(\frac{2}{3}\right)^n=2\cdot\frac{2/3}{1-2/3}=2\cdot2=4.$$
Hence the series converges with sum $4$.

(2) For $n\ge1$,
$$1-\frac{2}{n(n+1)}=\frac{(n-1)(n+2)}{n(n+1)},$$
so
$$\ln\!\left(1-\frac{2}{n(n+1)}\right)=\ln(n-1)+\ln(n+2)-\ln n-\ln(n+1).$$
Regroup as
$$\ln(n+2)-\ln(n+1)-\bigl(\ln n-\ln(n-1)\bigr).$$
The partial sum telescopes:
$$S_n=\ln(n+2)-\ln n-\ln 3.$$
As $n\to+\infty$, $\ln(n+2)-\ln n=\ln\!\left(1+\frac2n\right)\to0$, so $S_n\to-\ln3$.
The series converges with sum $-\ln 3$.

(3) Using $\arctan a+\arctan b=\arctan\frac{a+b}{1-ab}$ with $x=-n$ and $y=n+1$:
$$\arctan(-n)+\arctan(n+1)=\arctan\frac{1}{1+n(n+1)}=\arctan\!\left(\frac{1}{n^2+n+1}\right).$$
Since $\arctan(-n)=-\arctan(n)$,
$$\arctan\!\left(\frac{1}{n^2+n+1}\right)=\arctan(n+1)-\arctan(n).$$
The partial sum telescopes:
$$S_n=\arctan(n+1)-\arctan(1)=\arctan(n+1)-\frac{\pi}{4}.$$
As $n\to+\infty$, $\arctan(n+1)\to\pi/2$. Hence $S_n\to\frac{\pi}{4}$: the series converges with sum $\pi/4$.

Conclusion. (1) converges, sum $4$; (2) converges, sum $-\ln3$; (3) converges, sum $\pi/4$.