Question

Does this geometric series converge or diverge?

Determine whether the geometric series is convergent or divergent.
`oo`
`sum_(n=0) 1/(sqrt19)^n`

If it is convergent, find its sum.

Answer 1

Yes.

Sum `19/(19-sqrt(19))`

Explanation:

The first three terms are:

`1 , 1/sqrt(19), 1/19`

Common ratio:

`(1/sqrt(19))/1 = (1/(19))/(1/(sqrt19))= (sqrt(19))/19`

Sum of a geometric series:

`a((1-r^n)/(1-r))`

Where `a` is the first term.
`r` is the common ratio.

If `r<1`

Then:

`lim_(n->oo)(r^n)= 0`

Then:

`a((1-r^n)/(1-r))= a/(1-r)` ( converges to finite value)

If `r>1`

`lim_(n->oo)(r^n)=oo`

And:

`a((1-r^n)/(1-r))` (diverges )

`(sqrt(19))/19< 1`

So:

`sum_(n=0)^oo 1/((sqrt(19))^n` (converges)

`a((1-r^n)/(1-r))=((1-0)/(1-(sqrt(19)/19)))=19/(19-sqrt(19))`