Problem 5.

Note that $\frac {1}{n (n+2)(n+4)}=\frac {1}{4} (\frac {1}{n (n+2)}-\frac {1}{(n+2)(n+4)})$ Substituting $n=1, 3, 5, ...$ , we find that it is a telescoping sum. All terms other than the first, $\frac {1}{4}(\frac{1}{1 (1+2)})=\frac{1}{12}$ cancel out. Thus $\frac {1}{12}=0.08333...$ is the answer.

Clearly, we can see that the $nth$ term is given by the pattern:

${ a }_{ n }=\frac { 1 }{ (2n-1)(2n+1)(2n+3) }$

We need to calculate

$\quad \quad \sum _{ n=1 }^{ \infty }{ { a }_{ n } } \\ \Rightarrow \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ (2n-1)(2n+1)(2n+3) } }$

Since, there is no absolute pattern for such terms, so we need to simplify this sum onto simpler ones, and calculate them.

Now, by expanding $2$ terms out of the three terms alternatively, we can see that:

$(2n-1)(2n+1)\quad =\quad 4{ n }^{ 2 }-1\\ (2n+1)(2n+3)\quad =\quad 4{ n }^{ 2 }+8n+3\\ (2n-1)(2n+3)\quad =\quad { 4n }^{ 2 }+4n-3$

So, the required sum can be written as,

$\quad \quad \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ 8 } } \frac { (2n+1)(2n+3)+(2n-1)(2n+1)-2(2n-1)(2n+3) }{ (2n-1)(2n+1)(2n+3) } \\ \Rightarrow \frac { 1 }{ 8 } \sum _{ n=1 }^{ \infty }{ \frac { (2n+1)(2n+3) }{ (2n-1)(2n+1)(2n+3) } +\frac { (2n-1)(2n+1) }{ (2n-1)(2n+1)(2n+3) } -\frac { 2(2n-1)(2n+3) }{ (2n-1)(2n+1)(2n+3) } } \\ \Rightarrow \frac { 1 }{ 8 } \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ (2n-1) } +\frac { 1 }{ (2n+3) } -\frac { 2 }{ (2n+1) } }$

Now, by inserting the first few values, we can see the pattern, i.e,

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

Now, looking at the pattern, we can see that some all terms get cancelled, and the ones in the end will tend to xero, So, we are left with,

$=\frac { 1 }{ 8 } (1-\frac { 2 }{ 3 } +\frac { 1 }{ 3 } )\\ =\frac { 1 }{ 8 } \times \frac { 2 }{ 3 } \\ =\frac { 1 }{ 12 } =0.0833\quad$

CHEERS!:)