Question 388736
{{{sqrt(5)/sqrt(3)}}}{{{""+""}}}{{{sqrt(3)/sqrt(5)}}}{{{""+""}}}{{{expr(3/5)expr(sqrt(3)/sqrt(5))}}}
{{{"+..."}}}
<pre>
We check to see if it is a geometric series:

Dividing the second term by the first term:

{{{sqrt(3)/sqrt(5)}}}{{{"÷"}}}{{{sqrt(5)/sqrt(3)}}}{{{""=""}}}{{{sqrt(3)/sqrt(5)}}}{{{""*""}}}{{{sqrt(3)/sqrt(5)}}}{{{""=""}}}{{{3/5}}}

Dividing the third term by the second term:

{{{expr(3/5)expr(sqrt(3)/sqrt(5))}}}{{{"÷"}}}{{{sqrt(3)/sqrt(5)}}}{{{""=""}}}{{{expr(3/5)expr(sqrt(3)/sqrt(5))}}}{{{""*""}}}{{{sqrt(5)/sqrt(3)}}}{{{""=""}}}{{{expr(3/5)expr(cross(sqrt(3))/cross(sqrt(5)))}}}{{{""*""}}}{{{cross(sqrt(5))/cross(sqrt(3))}}}{{{""=""}}}{{{3/5}}}

Since we got the same thing both times, it's a geometric series
and the common ratio is what we got, namely {{{3/5}}}

The formula for the infinite sum is:

{{{S[infinity]}}}{{{""=""}}}{{{a[1]/(1-r)}}}

where {{{a[1]=sqrt(5)/sqrt(3)}}}

Let's rationalize the denominator of that

{{{a[1]=expr(sqrt(5)/sqrt(3))*expr(sqrt(3)/sqrt(3))=sqrt(15)/3}}}

and {{{r=3/5}}}. Substituting:

{{{S[infinity]}}}{{{""=""}}}{{{expr(sqrt(15)/3)/(1-3/5)}}}{{{""=""}}}{{{expr(sqrt(15)/3)/(2/5)}}}{{{""=""}}}{{{expr(sqrt(15)/3)*expr(5/2)}}}{{{""=""}}}{{{5sqrt(15)/6}}}

Edwin</pre>