How do you apply the ratio test to determine if $Sigma (135* * * (2n-1))/(n!)^2$ from $n=[1,oo)$ is convergent to divergent?

Question

How do you apply the ratio test to determine if $Sigma (135* * * (2n-1))/(n!)^2$ from $n=[1,oo)$ is convergent to divergent?

1 Answer

Impact of this question

2218 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-17T13:06:32

The series:

$sum_(n=1)^oo (prod_1^n 2k-1)/(n!)^2$

is convergent.

Explanation:

Write the series as:

$sum_(n=1)^oo (prod_1^n 2k-1)/(n!)^2$

The ratio test states that a sufficient condition for $sum_(n=1)^oo a_n$ to converge is that:

$L = lim_(n->oo) abs(a_(n+1)/a_n) < 1$

We then evaluate the ratio for the series at hand:

$abs(a_(n+1)/a_n) = frac ((prod_1^(n+1) (2k-1))/((n+1)!^2)) ((prod_1^n (2k-1))/(n!)^2) = ( (prod_1^(n+1) (2k-1)) / (prod_1^n (2k-1)) ) ((n!^2)/((n+1)!^2)) = (2(n+1)-1)/((n+1)^2) = (2n+1)/((n+1)^2)$

so that:

$lim_(n->oo) abs(a_(n+1)/a_n) = 0$

which proves the series is convergent.

Science

Math

Humanities

... and beyond

How do you apply the ratio test to determine if $Sigma (135* * * (2n-1))/(n!)^2$ from $n=[1,oo)$ is convergent to divergent?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you apply the ratio test to determine if Sigma (1*3*5* * * (2n-1))/(n!)^2Sigma (1*3*5* * * (2n-1))/(n!)^2 from n=[1,oo)n=[1,oo) is convergent to divergent?

1 Answer

Explanation:

Related questions

Impact of this question

How do you apply the ratio test to determine if $Sigma (135* * * (2n-1))/(n!)^2$ from $n=[1,oo)$ is convergent to divergent?