Convergence of n=1 to infinity[ n(n+2)/(n+3)^2 ]

[csmajor86]

How do I find if the series SUM[from n=1 to infinity] n(n+2)/(n+3)[sup:29rq1fym]2[/sup:29rq1fym] converges or diverges?

I don't even know where to start.

Thanks,
CS_Major

 

[skeeter]

use the most simple test , the nth term test for divergence ... note that the nth term of your stated series does not approach 0 as n -> infinity (as a matter of fact, it approaches 1)

 

[daon]

...

 

[skeeter]

Note that for $\displaystyle n \ge 1$:

$\displaystyle \frac{n^2 + 2n}{(n+3)^3} \ge \frac{n^2}{(n+3)^3} \ge \frac{n^2}{(n+3n)^3} \ge \frac{1}{64}\frac{1}{n}$

By the limit comparison test,

$\displaystyle \sum \frac{n^2 + 2n}{(n+3)^3} \,\, \ge \,\, \frac{1}{64} \sum \frac{1}{n} \,\, \rightarrow \,\, \infty$

So it must diverge.

note that in the original problem, the denominator is squared, not cubed.

 

[daon]

Thats funny, I could have sworn it was cubed. Oh well.