Finding the limit of a series

Question

I got the n'th term but clueless about what to do next

Answer 1

We have $$t_r = \dfrac{r(r+1)(2r+1)/6}{r^2(r+1)^2/4} = \dfrac23 \dfrac{2r+1}{r(r+1)} = \dfrac23 \left(\dfrac1r + \dfrac1{r+1}\right)$$ Hence, \begin{align} S_n & = \sum_{r=1}^n (-1)^r t_r = \sum_{r=1}^n (-1)^r \dfrac23 \left(\dfrac1r + \dfrac1{r+1}\right) = \dfrac23 \left(\sum_{r=1}^n \dfrac{(-1)^r}r + \sum_{r=1}^n \dfrac{(-1)^r}{r+1} \right)\\ & = \dfrac23 \left(-\dfrac11 + \sum_{r=2}^n \dfrac{(-1)^r}r + \sum_{r=1}^{n-1} \dfrac{(-1)^r}{r+1} + \dfrac{(-1)^n}{n+1} \right)\\ & = \dfrac23 \left(-1 + \sum_{r=1}^{n-1} \dfrac{(-1)^{r+1}}{r+1} + \sum_{r=1}^{n-1} \dfrac{(-1)^r}{r+1} + \dfrac{(-1)^n}{n+1}\right)\\ & = -\dfrac23 \left(1+ \dfrac{(-1)^{n+1}}{n+1}\right) \end{align} Now conclude the limit of $S_n$.

Answer 2

HINT:

$$t_r=\frac23\cdot\frac{2r+1}{r(r+1)}=\frac23\cdot\frac{r+(r+1)}{r(r+1)}$$

$$=\frac23\left(\frac1r+\frac1{r+1}\right)$$

Set a few values of $r$ to find the Telescoping nature

and for integer $m>0,$ $$\frac32S_m=-1+\frac{(-1)^m}{m+1}$$