# Thread: Whether a series converges or diverges.

1. ## Whether a series converges or diverges.

I am given the series n^2/(n^3 + 1) and I don't know how to put the Riemann sum symbol in front of it (assume its there). My guess is that the series diverges because -ve terms produce a -ve result because of the n^3 in the denominator and all +ve terms produce +ve results (partial sums are increasing?)

Once I have the suspicion that the series diverges do I perform the integral test? Or am I lost and the series converges?

2. On a second attempt I got that while there is a "seeming" oscillating pattern, the series is bounded above when there is a number M such that An is smaller or equal to M for all n are greater or equal to 1. Also, since the partial sums for "n is greater or equal to 1" are increasing, the series converges by the monotonic theorem.

3. Originally Posted by Undefdisfigure
I am given the series n^2/(n^3 + 1) and I don't know how to put the Riemann sum symbol in front of it (assume its there). My guess is that the series diverges because -ve terms produce a -ve result because of the n^3 in the denominator and all +ve terms produce +ve results (partial sums are increasing?)

Once I have the suspicion that the series diverges do I perform the integral test? Or am I lost and the series converges?
$\displaystyle \sum_{n=1}^\infty \frac{n^2}{n^3 + 1}$ diverges by the comparison test.

$\displaystyle \frac{n^2}{n^3 + 1} \geq \frac{n^2}{n^3 + n^3} = \frac{1}{2}\left( \frac{1}{n} \right)$ and $\displaystyle \sum_{n=1}^\infty \frac{1}{n}$ is well known to be divergent.

4. Originally Posted by mr fantastic
$\displaystyle \sum_{n=1}^\infty \frac{n^2}{n^3 + 1}$ diverges by the comparison test.

$\displaystyle \frac{n^2}{n^3 + 1} \geq \frac{n^2}{n^3 + n^3} = \frac{1}{2}\left( \frac{1}{n} \right)$ and $\displaystyle \sum_{n=1}^\infty \frac{1}{n}$ is well known to be divergent.

Umm, could you use another method please? This method is valid for section 12.4 of my book where the comparison method is introduced. I'm in section 12.3 and they don't expect me to use the comparison method yet. Unless you can prove me wrong in some other way, I will say that the series converges by the monotonic theorem.

5. Originally Posted by Undefdisfigure
I am given the series n^2/(n^3 + 1) and I don't know how to put the Riemann sum symbol in front of it (assume its there). My guess is that the series diverges because -ve terms produce a -ve result because of the n^3 in the denominator and all +ve terms produce +ve results (partial sums are increasing?)
there are no negative terms for this series, and just because the partial sums are increasing does not mean the series diverge. the first few partial sums of $\displaystyle \sum \frac 1{n^2}$ increase for instance

Once I have the suspicion that the series diverges do I perform the integral test? Or am I lost and the series converges?
the integral test is not one you want to use unless necessary, it is usually a pain. also, certain conditions must be fulfilled for you to use it (incidentally, you can use it here)

Originally Posted by Undefdisfigure
Umm, could you use another method please? This method is valid for section 12.4 of my book where the comparison method is introduced. I'm in section 12.3 and they don't expect me to use the comparison method yet.
well, in that case, go with the integral test, since i assumed you've learned it already.

Unless you can prove me wrong in some other way, I will say that the series converges by the monotonic theorem.
what theorem are you referring to? are you sure the theorem is not for sequences as opposed to series? the series diverges, so you are definitely applying the theorem wrong

6. Originally Posted by Undefdisfigure
Umm, could you use another method please? This method is valid for section 12.4 of my book where the comparison method is introduced. I'm in section 12.3 and they don't expect me to use the comparison method yet. Unless you can prove me wrong in some other way, I will say that the series converges by the monotonic theorem.
Well, the series diverges. That's an established mathematical fact. So good luck proving otherwise!

But before embarking on what will prove to be a futile effort (provided the mathematics is done correctly all the way), you might want to consider the following:

Integral comparison test:

$\displaystyle \sum_{n=1}^\infty \frac{n^2}{n^3 + 1} = \frac{1^2}{1^3 + 1} + \sum_{n=2}^\infty \frac{n^2}{n^3 + 1} = \frac{1}{2} + \sum_{n=2}^\infty \frac{n^2}{n^3 + 1} > \frac{1}{2} + \int_2^{+ \infty} \frac{x^2}{x^3 + 1} \, dx$.

$\displaystyle \int_2^{+ \infty} \frac{x^2}{x^3 + 1} \, dx = \lim_{\alpha \rightarrow + \infty} \int_2^\alpha \frac{x^2}{x^3 + 1} \, dx = \frac{1}{3} \lim_{\alpha \rightarrow + \infty} \left[ \ln (x^3 + 1) \right]_2^\alpha = +\infty$.

So, unsurprisingly by now I'd hope, the series diverges.

7. Hey Jhevon, your from Jamaica, awesome!