Question 616172
 5+5+(5/2)+(2/6)+(5/24)+...
<pre>
Those denominators are factorials:

4! = 4·3·2·1 = 24
3! = 3·2·1 = 6
2! = 2·1 = 2
1! = 1

also 0! is defined as 1.

Therefore the series:

 5 + 5 + {{{5/2}}} + {{{5/6}}} + {{{5/24}}} + ...

is really

 {{{5/1}}} + {{{5/1}}} + {{{5/2}}} + {{{5/6}}} + {{{5/24}}} + ...

which is really:

 {{{5/0!}}} + {{{5/1!}}} + {{{5/2!}}} + {{{5/3!}}} + {{{5/4!}}} + ... + {{{5/n!}}} + ...

So, the nth term is  {{{5/n!}}}, where we start n with 0.

We put a <font face="symbol">S</font> in front of that with n=0 on the

bottom and {{{infinity}}} on the top, since the ellipsis "..." with
nothing after it indicates that it never ends.  So the sigma notation
is

{{{sum((5/n!),n=0,infinity)}}}

By the way, you can use other letters for the "index" (or "dummy variable")
besides n.

These answers are just as good:

{{{sum((5/k!),k=0,infinity)}}}, {{{sum((5/i!),i=0,infinity)}}}, {{{sum((5/j!),j=0,infinity)}}}, {{{sum((5/m!),m=0,infinity)}}}, etc.

You can also start the index (or dummy variable) at 1 by subtracting 1
from the dummy variable after the sigma:

{{{sum((5/n!),n=0,infinity)}}} is the same as {{{sum((5/(n-1)!),n=1,infinity)}}}

In fact you can add any integer to the start of the dummy variable and subtract
the same integer from the dummy variable after the sigma.  For instance,  

{{{sum((5/n!),n=0,infinity)}}} is the same as {{{sum((5/(n-7)!),n=7,infinity)}}}
number 


Edwin</pre>