vazelinahS

2020-11-08

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 sum ${S}_{n}$ of the indinite 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.
$\sum _{k=1}^{\mathrm{\infty }}\frac{2}{\left(2k-1\right)\left(2k+1\right)}$

Elberte

Step 1.
We have to consider the infinite series:
$\sum _{k=1}^{\mathrm{\infty }}\frac{2}{\left(2k-1\right)\left(2k+1\right)}$
(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
$\sum _{k=1}^{\mathrm{\infty }}\frac{2}{\left(2k-1\right)\left(2k+1\right)}$
(a) To find the first four partial sums ${S}_{1},{S}_{2},{S}_{3},{S}_{4}$ of the series.
${S}_{1}={a}_{1}=\frac{2}{1\left(3\right)}=\frac{2}{3}$
${S}_{2}={S}_{1}+{a}_{2}=\frac{2}{3}+\frac{2}{3.5}=\frac{2}{3}+\frac{2}{15}=\frac{4}{5}$
${S}_{3}={S}_{2}+{a}_{3}=\frac{4}{5}+\frac{2}{5.7}=\frac{4}{5}+\frac{2}{35}=\frac{6}{7}$
${S}_{4}={S}_{3}+{a}_{4}=\frac{6}{7}+\frac{2}{7.9}=\frac{6}{7}+\frac{2}{63}=\frac{8}{9}$
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.
$\sum _{k=1}^{\mathrm{\infty }}\frac{2}{\left(2k-1\right)\left(2k+1\right)}$
We can also write $\sum _{k=1}^{\mathrm{\infty }}\frac{2}{\left(2k-1\right)\left(2k+1\right)}$ as $\sum _{k=1}^{\mathrm{\infty }}\frac{2}{\left(4{k}^{2}-1\right)}$
$\sum _{k=1}^{\mathrm{\infty }}\frac{2}{\left(4{k}^{2}-1\right)}=\sum _{k=1}^{\mathrm{\infty }}\left(-\frac{1}{2\left(k+\frac{1}{2}\right)}+\frac{1}{2\left(k-\frac{1}{2}\right)}\right)$
$=\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{9}\right)+...$
On cancelling out the terms we only left wiyh 1, So
$\sum _{k=1}^{\mathrm{\infty }}\frac{2}{\left(2k-1\right)\left(2k+1\right)}=1$
Therefore, the value of the series is 1.

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