Three convergent series with explicit sums

(1) Geometric series. 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. We recognize
$$\sum_{n\ge1}\frac{2^{n+1}}{3^n}=2\sum_{n\ge1}\left(\frac{2}{3}\right)^n,$$
a geometric series with ratio $\frac23$. Its partial sums are
$$S_n=2\sum_{k=1}^n\left(\frac{2}{3}\right)^k=2\cdot\frac{2}{3}\cdot\frac{1-(2/3)^n}{1-2/3}.$$
Taking $n\to+\infty$:
$$S=2\cdot\frac{2/3}{1/3}=2\cdot 2=4.$$

(2) Logarithmic telescoping series. For $n\ge1$,
$$1-\frac{2}{n(n+1)}=\frac{(n-1)(n+2)}{n(n+1)},$$
hence
$$\ln\!\left(1-\frac{2}{n(n+1)}\right)=\ln(n-1)+\ln(n+2)-\ln n-\ln(n+1).$$
Regroup the terms as
$$\ln(n+2)-\ln(n+1)-\bigl(\ln n-\ln(n-1)\bigr).$$
The partial sum of order $n$ telescopes:
$$S_n=\ln(n+2)-\ln n-\ln 3=\ln\!\left(1+\frac2n\right)-\ln 3.$$
Since $\ln(1+2/n)\to0$, we get $S_n\to-\ln 3$. The series converges with sum $-\ln 3$.

(3) Arctangent telescoping series. Using $\arctan x+\arctan y=\arctan\frac{x+y}{1-xy}$ 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 $n\to+\infty$, $\arctan(n+1)\to\frac{\pi}{2}$, so $S_n\to\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}$. The series converges with sum $\frac{\pi}{4}$.