Answer
Converges to $1$.
Work Step by Step
Consider $\Sigma_{n=1}^{\infty} \dfrac{1}{k (k+1)}=\Sigma_{n=1}^{\infty} \dfrac{1}{k }-\dfrac{1}{{(k+1)}}$
$\implies s_n=1-\dfrac{1}{n+1}$
$\implies \lim\limits_{n \to \infty} s_n= \lim\limits_{n \to \infty} (1-\dfrac{1}{n+1}) \\ \implies \lim\limits_{n \to \infty} s_n=1$
Further, $T_{2n+1}=T_{2n}+\dfrac{1}{n+1} \\ = (1-\dfrac{1}{n+1})+\dfrac{1}{n+1} \\=1$
Now, $\lim\limits_{n \to \infty} s_n= \lim\limits_{n \to \infty} (1-\dfrac{1}{n+1}) \\=1$
This means that both series are convergent and they converge to $1$.
