Sum $\rightarrow$ Product

Given : $1 - \dfrac{1}{3} + \dfrac{1}{5} - \dfrac{1}{7} + \dfrac{1}{9} - \dfrac{1}{11} + ... = \dfrac{\pi}{4}$

Now, just take LCM for every two consecutive fractions. $\begin{array}{c}~\implies \dfrac{3 - 1}{1 \times 3} + \dfrac{7 - 5}{5 \times 7} + \dfrac{11 - 9}{9 \times 11} + ... = \dfrac{\pi}{4} \\ \implies 2 \left(\dfrac{1}{1 \times 3} + \dfrac{1}{5 \times 7} + \dfrac{1}{9 \times 11} + ... \right) = \dfrac{\pi}{4} \\ \implies \dfrac{1}{1 \times 3} + \dfrac{1}{5 \times 7} + \dfrac{1}{9 \times 11} + ... = \dfrac{\pi}{8} = \dfrac{\pi}{\color{#3D99F6}x} \\ \end{array}$ $\therefore \color{#3D99F6}x = 8$

Similar solution with @Ram Mohith 's

$\begin{aligned} {\color{#3D99F6}1 - \frac 13} + {\color{#D61F06}\frac 15 - \frac 17} + {\color{#3D99F6}\frac 19 - \frac 1{11}} + \cdots & = \frac \pi 4 & \small \color{#3D99F6} \text{Given} \\ {\color{#3D99F6} \frac {3-1}{1\times 3}} + {\color{#D61F06} \frac {7-5}{5\times 7}} + {\color{#3D99F6} \frac {11-9}{9\times 11}} + \cdots & = \frac \pi 4 \\ {\color{#3D99F6} \frac 2{1\times 3}} + {\color{#D61F06} \frac 2{5\times 7}} + {\color{#3D99F6} \frac 2{9\times 11}} + \cdots & = \frac \pi 4 & \small \color{#3D99F6} \text{Divide both sides by }2 \\ \implies \frac 1{1\times 3} + \frac 1{5\times 7} + \frac 1{9\times 11} + \cdots & = \frac \pi {\boxed 8} \end{aligned}$

Therefore, $x = \boxed 8$ .

Observe that $\begin{aligned}\dfrac{1}{1+x^2} = 1 -x^2 +x^4 -x^6 +\cdots \qquad |x| < 1 \end{aligned}$ Integrating we obtain $\begin{aligned} \int \dfrac{1}{1+x^2}& = x- \dfrac{x^3}{3} + \dfrac{x^5}{5} -\dfrac{x^7}{7}+\dfrac{x^9}{9} -\dfrac{x^{11}}{11}+\cdots \\ \text{arc}\tan x & = \dfrac{x\,(3-x^2)}{3}+ \dfrac{x^5\,(7-5x^2)}{3\times 5} + \dfrac{x^9\,(11-9x^2)}{9\times 11}+ \cdots \\ \therefore \int_{0}^{1}\text{arc}\tan x & = \dfrac{\pi}{4}= \dfrac{2}{3} +\dfrac{2}{3\times 5} +\dfrac{2}{7\times 9}+\dfrac{2}{9\times 11}+\cdots \\ \dfrac{\pi}{8} &= \dfrac{1}{1 \times 3} + \dfrac{1}{5 \times 7} + \dfrac{1}{9 \times 11} +\cdots\end{aligned}$ Hence $x=8$ .

$\dfrac{1}{a(a+2)}=\dfrac{1}{2}\left(\dfrac{1}{a}-\dfrac{1}{a+2}\right)$

Thus the result is $\dfrac{1}{2}\dfrac{\pi}{4}$ hence $x=2\times 4=\boxed{8}$ .

Related wiki: Partial Fractions - Linear Factors

$\dfrac {1}{n\times(n+2)} = \dfrac{A}{n} + \dfrac {B}{n+2} = \dfrac{n(A+B)+2A}{n\times(n+2)}$

Solve for $A$ and $B$

$A=\dfrac{1}{2}, B= -\dfrac{1}{2}$

Then,

$\dfrac{1}{1\times3} + \dfrac{1}{5\times7} + \dfrac{1}{9\times11} + \cdots = \dfrac{1}{2} \left( \dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}-\dfrac{1}{11}+\cdots \right) = \dfrac{\pi}{\color{#3D99F6}8}$

Therefore, $x=\boxed {8}$

So,we have

$1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\dots=\frac{\pi}{4}=a$ & $\frac{1}{1 \times 3}+\frac{1}{7 \times 5}+\frac{1}{11 \times 9}+\dots=b$

which is

$\frac{3-1}{3 \times 1}+\frac{7-5}{7 \times 5}+\frac{11-9}{11 \times 9}+\dots=\frac{\pi}{4} \implies 2b$ .So, $\frac{\pi}{8}=b$

Hence, $\boxed{\large{x=8}}$

Group two consecutive terms in the first series and show half of this is equal to to each term in the second series. Therefore,

Answer=π/8

First note that

$a-b=(\dfrac{1}{b}-\dfrac{1}{a})(ab)$

Therefore,

$\frac{1}{1}-\frac{1}{3} = 2(\frac{1}{1 \times 3}),$ $\frac{1}{5}-\frac{1}{7} = 2(\frac{1}{5 \times 7}),$ $\frac{1}{9}-\frac{1}{11} = 2(\frac{1}{9 \times 1}),$ $...$

Finally,

$\frac{1}{1 \times 3} + \frac{1}{5 \times 7} + \frac{1}{9 \times 11} + ... = \frac{1}{2}(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+...) = \frac{\pi}{8}$

$\therefore x = \boxed{8}$

Let us express the first sum as:

$S_1 \ = \ 1 \ - \ \dfrac{1}{3} \ + \ \dfrac{1}{5} \ - \ \dfrac{1}{7} \ + \ … \ \Rightarrow \ \displaystyle\sum_{n=0}^{\infty} \ \dfrac{1}{4n \ + \ 1} \ - \ \dfrac{1}{4n \ + \ 3}$

We can express the second sum as:

$S_2 \ = \ \dfrac{1}{1 \ \times \ 3} \ + \ \dfrac{1}{5 \ \times \ 7} \ + \ \dfrac{1}{9 \ \times \ 11} \ + \ … \ \Rightarrow \ \displaystyle\sum_{n=0}^{\infty} \ \dfrac{1}{(4n \ + \ 1)(4n \ + \ 3)}$

Since we have similar terms in the denominators of the expressions in the sums, we can simply find partial fractions for the expression enclosed in the second sum, to make it look like the first sum. Let us set:

$\dfrac{A}{4n \ + \ 1} \ + \ \dfrac{B}{4n \ + \ 3} \ = \ \dfrac{1}{(4n \ + \ 1)(4n \ + \ 3)}$

We now must solve for $A$ and $B$ . Multiplying both sides by $(4n \ + \ 1)(4n \ + \ 3)$ gives us:

$A(4n \ + \ 3) \ + \ B(4n \ + \ 1) \ = \ 1 \ \Rightarrow \ 4n(A \ + \ B) \ + \ 3A \ + \ B \ = \ 1$

The sums in question are $\dfrac{\pi}{4}$ and $\dfrac{\pi}{x}$ , so it is just a matter of picking a constant to multiply the first sum by, to yield the solution for $x$ and the second sum. Because of this, we must get rid of $n$ by simply setting $A \ + \ B \ = \ 0$ , giving us two equations: $A \ + \ B \ = \ 0$ and $3A \ + \ B \ = \ 1$ . Solving this system gives us $A \ = \ \frac{1}{2}$ and $B \ = \ -\frac{1}{2}$ . We can then rewrite the second sum with $A$ and $B$ :

$S_2 \ = \ \displaystyle\sum_{n=0}^{\infty} \ \dfrac{1}{2(4n \ + \ 1)} \ - \ \dfrac{1}{2(4n \ + \ 3)} \ = \ \dfrac{1}{2} \ \displaystyle\sum_{n=0}^{\infty} \ \dfrac{1}{4n \ + \ 1} \ - \ \dfrac{1}{4n \ + \ 3} \ = \ \dfrac{1}{2} S_1 \ = \ \dfrac{\pi}{8}$

Therefore, $x \ = \ 8$ .

Use excell to load the fractions and let it calculate the tendency. Obviously: The tendency is 8!!!!!!!!

After you see the pattern $\frac{1}{n}-\frac{1}{n+2} = \frac{n+2}{n(n+2)}-\frac{n}{n(n+2)} = \frac{2}{n(n+2)}$ you can conclude by pairing every two terms in the first series, its value is double the value of second series and thus x=8.

Let $1 - \dfrac{1}{3} + \dfrac{1}{5} - \dfrac{1}{7} + \dfrac{1}{9} - \dfrac{1}{11} + ...= \dfrac{\pi}{4}$ and $S = \sum_{n = 0}^{\infty} \dfrac{1}{(4n + 1)(4n + 3)}$ .

Using partial fractions we obtain:

$\dfrac{1}{(4n + 1)(4n + 3)} = \dfrac{A}{4n + 1} + \dfrac{B}{4 n + 3} \implies 1 = 4(A + B)n + 3A + B = 1 \implies$

$A + B = 0$

$3A + B = 1$

$\implies A = \dfrac{1}{2}$ and $B = -\dfrac{1}{2}$

$\implies S = \dfrac{1}{2}\sum_{n = 0}^{\infty} (\dfrac{1}{4n + 1} - \dfrac{1}{4n + 3}) = \dfrac{1}{2}(1 - \dfrac{1}{3} + \dfrac{1}{5} - \dfrac{1}{7} + \dfrac{1}{9} - \dfrac{1}{11} + ... ) = \dfrac{1}{2}(\dfrac{\pi}{4}) = \dfrac{\pi}{8} \implies \boxed{x = 8}$ .

Given : 1 - 1/3 + 1/5 - 1/7 + 1/9 + ..... = $pi$ /4

So, we can group this as follows (1 + 1/5 + 1/9 + 1/13) - (1/3 + 1/7 + 1/11 + 1/15) = $pi$ /4

This gives, sum of 1/(4n+1) - sum of 1/(4n+3) = $pi$ /4 , for n = 0 to infinity

Hence, sum of 1/(4n+1) - 1/(4n+3) = $pi$ /4

Taking lcm, sum of (4n+3-4n-1)/{(4n+1)(4n+3)} = $pi$ /4, which implies sum of 2/{(4n+1)(4n+3)} = $pi$ /4

Now taking the second equation,

1/(1 3) + 1/(5 7) + 1/(9*11) + .........

can be written as sum of 1/{(4n+1)(4n+3) for n = 0 to infinity

Hence, this will be equal to $pi$ /8

So x = 8