# Thread: Infinite Positive Series Convergence

1. ## Infinite Positive Series Convergence

does series (n= 1 --> inf) sqrt(n)/(4n + 9) converge or diverge?

I said that it does converge, and I used the comparison test with b(sub n) = n/(4n + 9) got the limit at (1/4) for b(sub n). But the back of the book said I was wrong, what did I do wrong?

my steps: 0 < a(sub n) < b(sub n)

b(sub n) converges as limit approaches infinity, so doesn't that mean a(sub n) converges?

Thanks for any input. Sorry I did not use latex on this.

2. We have:

$\displaystyle\lim_{n \to{+}\infty}{\dfrac{\sqrt{n}}{4n+9}:\dfrac{1}{\sq rt{n}}}=\dfrac{1}{4}\neq 0$

and

$\displaystyle\sum_{n=1}^{+\infty}\dfrac{1}{n^{1/2}}$

is divergent, so the given series is also divergent.

Fernando Revilla

3. You may be confusing the sequence converging with the series converging.

In order that the series $\sum_{n=0}^\infty a_n$ converge to any number, it is necessary (but not sufficient) that the sequence $\{a_n\}$ converge to 0.

4. Originally Posted by HallsofIvy
You may be confusing the sequence converging with the series converging.

In order that the series $\sum_{n=0}^\infty a_n$ converge to any number, it is necessary (but not sufficient) that the sequence $\{a_n\}$ converge to 0.
Certainly your post responds better than mine to Warrenx's question. I didn't pay attention.

Fernando Revilla