Consider the following infinite series. a.Find the first four partial sums S_1,S_2,S_3, and S_4 of the series. b.Find a formula for the nth partial su

Step 1.
We have to consider the infinite series:
$\sum _{k=1}^{\mathrm{\infty }}\frac{2}{(2k-1)(2k+1)}$
(a) To find the first four partial sums $S_1,S_2,S_3,S_4$ of the series.
b.Find a formula or the nth partial sum $S_n$ of the infinite series.Use this formula to find the next four partial sums $S_5,S_6,S_7$ and $S_8$ of the infinite series.
c. Make a conjecture for the value of the series
Step 2
$\sum _{k=1}^{\mathrm{\infty }}\frac{2}{(2k-1)(2k+1)}$
(a) To find the first four partial sums $S_1,S_2,S_3,S_4$ of the series.
$S_1=a_1=\frac{2}{1(3)}=\frac{2}{3}$
$S_2=S_1+a_2=\frac{2}{3}+\frac{2}{3.5}=\frac{2}{3}+\frac{2}{15}=\frac{4}{5}$
$S_3=S_2+a_3=\frac{4}{5}+\frac{2}{5.7}=\frac{4}{5}+\frac{2}{35}=\frac{6}{7}$
$S_4=S_3+a_4=\frac{6}{7}+\frac{2}{7.9}=\frac{6}{7}+\frac{2}{63}=\frac{8}{9}$
Step 3
(b) From the partial sums $S_1,S_2,S_3,S_4$ we see that the numerator is 2n and the denominator is 2n+1
So, we can write
$S_n=\frac{2n}{2n+1}$
Now we find $S_5,S_6,S_7,S_8$
$S_n=\frac{2n}{2n+1}$
$S_5=\frac{10}{11}$
$S_6=\frac{12}{13}$
$S_7=\frac{14}{15}$
$S_8=\frac{16}{17}$
(c) To make a conjecture of the value of the series.
$\sum _{k=1}^{\mathrm{\infty }}\frac{2}{(2k-1)(2k+1)}$
We can also write $\sum _{k=1}^{\mathrm{\infty }}\frac{2}{(2k-1)(2k+1)}$ as $\sum _{k=1}^{\mathrm{\infty }}\frac{2}{(4k^2-1)}$
$\sum _{k=1}^{\mathrm{\infty }}\frac{2}{(4k^2-1)}=\sum _{k=1}^{\mathrm{\infty }}(-\frac{1}{2(k+\frac{1}{2})}+\frac{1}{2(k-\frac{1}{2})})$
$=(1-\frac{1}{3})+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{7})+(\frac{1}{7}-\frac{1}{9})+...$
On cancelling out the terms we only left wiyh 1, So
$\sum _{k=1}^{\mathrm{\infty }}\frac{2}{(2k-1)(2k+1)}=1$
Therefore, the value of the series is 1.