Partial Fractions With A Side Of Pi

Notice the first equation can be rewritten as:

$\frac {2}{3} + \frac {2}{35} + \frac {2}{99} + \ldots = \frac {\pi}{4}$ by subtracting every two terms $(1 - \frac {1}{3} , \frac {1}{5} - \frac {1}{7} , \ldots$ ).

Factor out 2 and we will get 2 times the expression on the left side of the second equation. Divide both sides by two and we get $\frac {\pi}{8}$ . Thus, n = 8.

Let us consider the first two terms of the first series:

$1 - \dfrac {1}{3} = \dfrac {1}{1} - \dfrac {1}{3} = \dfrac {2}{1\times 3}$

The first two terms is twice of the first term of second series.

Similarly, the third and fourth terms of first series:

$\dfrac {1}{5} - \dfrac {1}{7} = \dfrac {2}{5\times 7}$

is again twice that of the second term of second series.

By induction, the first series is twice that of the second, hence the second series $= \dfrac {\pi}{8}\quad \Rightarrow n = \boxed {8}$

I don't know math. But I use programming and gives me an approximate value of n = 8.000127325980861 up to 10000th term of the series.:)

$1 - \frac {1}{3} + \frac {1}{5} - \frac {1}{7} + \ldots = \frac {\pi}{4}$

$\frac {1}{1 \cdot 3} + \frac {1}{5 \cdot 7} + \frac {1}{9 \cdot 11} + \ldots = \frac {\pi}{n}$

First put brackets around every two term set in the first equation

$(1 - \frac {1}{3}) + (\frac {1}{5} - \frac {1}{7}) + \ldots = \frac {\pi}{4}$

Then multiply out the brackets to get

$(\frac {2}{1 \cdot 3}) + (\frac {2}{5 \cdot 7}) + \ldots = \frac {\pi}{4}$

Notice that the first equation now looks similar to the second one. In fact the denominators are exactly the same but the numerators are twice that of the second equation so we divide by $2$ to get

$\frac {1}{1 \cdot 3} + \frac {1}{5 \cdot 7} + \ldots = \frac {\pi}{4} \div 2$

So now we have the answer to the second equation but it's not simplified so let's do that

$\frac {\pi}{4} \div 2 = \frac {\pi}{4} \cdot \frac {1}{2} = \frac {\pi}{8}$

That means that

$\frac {\pi}{n} = \frac {\pi}{8}$

Now multiply by $8$ to get

$\frac {8\pi}{n} = \pi$

Next multiply by $n$ to get

$8\pi = n\pi$

Now finally divide by $\pi$ to get the value of $n$

$n = 8$

$\frac{1}{1 \times 3} = \frac{1}{2} \times (1-\frac{1}{3})$ same for all the other pairs so answer is $\frac{1}{2} \times \frac{\pi}{4} = \frac{\pi}{8}$ where $n=\boxed{8}$ .

Simplifying in pairs, you see that the top equation is double that of the bottom equation. Therefore, n = 8.

Since $\frac { 1 }{ x(x+2) } =\frac { 1 }{ 2 } (\frac { 1 }{ x } +\frac { 1 }{ x+2 } )$ we can use this partial fraction decomposition to rewrite $\frac { 1 }{ 1\times 3 } +\frac { 1 }{ 5\times 7 } +\frac { 1 }{ 9\times 11 } +\cdots$ as $\frac { 1 }{ 2 } \times \left( 1-\frac { 1 }{ 3 } +\frac { 1 }{ 5 } -\frac { 1 }{ 7 } +\cdots \right)$ . Dividing the two expressions given to us in the original problem and solving for n, we find that n = 8.

Note that the first equation can be written as 1/2( 1/1 + 1/3 + 1/5 + ... ) = π/4 while the second equation can be written as 1/1 + 1/3 + 1/5 + ... = π/n. Thus multiply both sides of first equation by 2 to get ( 1/1 + 1/3 + 1/5 + ... ) = π/8 => n = 8

As $\dfrac{1}{1 \times 3}+\dfrac{1}{5 \times 7}+\dfrac{1}{9 \times 11}+ \cdots = \dfrac{\pi}{n}$ is a multiple of $1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+ \cdots = \dfrac{\pi}{4}$

one can rewrite that as

$a(\dfrac{1}{1 \times 3}+\dfrac{1}{5 \times 7}+\dfrac{1}{9 \times 11}+ \cdots) = 1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+ \cdots = \dfrac{\pi}{4}$

The idea is that every fraction of the LHS, of the form $\dfrac{a}{x \times (x+2)}$ had been decomposed into $\dfrac{1}{x}-\dfrac{1}{x+2}$

So,

$\dfrac{a}{x \times (x+2)}=\dfrac{1}{x}-\dfrac{1}{x+2}$

Applying the Cover-Up-Rule, so here taking both sides of the equation $*(x+2)$ and setting x to -2, the consequence is

$\dfrac{a}{-2}=-1$ , $a=2$ .

Hence, the limit of the second series is half as big as the one from the first series, or $n = 8$ .

1 - 1/3 = 2/(1X3) 1/5 - 1/7 = 2/(5 X 7) 1/9 - 1/11 = 2(9 X 11) .... .... Adding Pi/4 = 2 X Pi/n Then n = 8

simply

for(int i=1;i<=1000;i+=4)

{ val+= 1/(double)((i)*(i+2)) ;

}
cout<<val<<endl;

val=M_PI/val; cout<<val;