Question

How do you test the series `Sigma n^2/2^n` from n is `[0,oo)` for convergence?

Answer 1

the series converges

Explanation:

We can apply d'Alembert's ratio test:

Suppose that;

`S=sum_(r=1)^oo a_n \ \ `, and `\ \ L=lim_(n rarr oo) |a_(n+1)/a_n|`

Then

if L < 1 then the series converges absolutely;
if L > 1 then the series is divergent;
if L = 1 or the limit fails to exist the test is inconclusive.

So our series is;

`S = sum_(n=0)^oo n^2/2^n`

So our test limit is:

`L = lim_(n rarr oo) | ((n+1)^2/2^(n+1)) / {n^2/2^n} |`
`\ \ \ = lim_(n rarr oo) | (n+1)^2/2^(n+1) * 2^n/{n^2} |`
`\ \ \ = lim_(n rarr oo) | (n+1)^2/(2*2^n) * 2^n/{n^2} |`
`\ \ \ = lim_(n rarr oo) | (n+1)^2/2 * 1/{n^2} |`
`\ \ \ = lim_(n rarr oo) | 1/2 * ((n+1)/n)^2 |`
`\ \ \ = lim_(n rarr oo) | 1/2 * (1+1/n)^2 |`
`\ \ \ = | 1/2 * (1+0)^2 |`
`\ \ \ = 1/2`

and we can conclude that the series converges

Answer 2

This series converges

Explanation:

The easiest shot is the ratio test , so we look at:

`abs((a_(n+1))/(a_n))_(n to oo`

Here that means:

`abs(((n+1)^2/(2^(n+1)))/(n^2/(2^n)))_(n to oo)`

`= abs(((n+1)^2)/(2 n^2))_(n to oo)`

`= abs((1 + 2/n + 1/n^2)/(2 ))_(n to oo) = 1/2`

So `abs((a_(n+1))/(a_n))_(n to oo) < 1` which means that it converges absolutely.