Question #41dc8

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Answer 1 · 2017-11-14T13:43:54

See below.

This is a geometric series. The sum to infinity of a geometric series can be expressed as:

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

Where $a$ is the first term, $r$ is the common ratio and n is the nth term.

The limit to infinity of:

$lim_(n->oo)(a-r^n)=a$ if and only if $-1 < r < 1$

( since $r^n-> 0$ , this is convergence)

If $color(white)(88)1 < r < -1$

Then:

$r > 1color(white)(88)$

$lim_(n->oo)(a-r^n)=-oo$ ( this is divergence )

( for $r < -1$ the limit is undefined )

So from example:

Common difference is:

$(1/3)/(-1/sqrt(3))=(-1/(3sqrt(3)))/(1/3)=(-sqrt(3))/3$

$-1 < (-sqrt(3))/3 < 1$ ( so this is a convergent series )

Sum to infinity:

$-1/sqrt(3)(1/(1-(-sqrt(3))/3))=((-1)/sqrt(3))/(1+(sqrt(3))/3)=-sqrt(3)/(3+sqrt(3))$