In Series and Sequences, what is the difference between $a_n$ and $s_n$?

Question

I'm having a lot of trouble on this topic; I know the test for divergence is that if limit as n goes to infinity is not zero, then it diverges. That was the answer for "determine C/D for $$\sum_{i=0}^\infty \frac{4n^2-n^3}{10+2n^3}$$ It uses An notation (from http://tutorial.math.lamar.edu/Classes/CalcII/ConvergenceOfSeries.aspx)

But then I did a problem that was "determine con/div of: $$\sum_{i=0}^\infty \frac{1}{n^2+3n+2} $$ and it converges, but its limit is zero as n goes to infinity. (see http://tutorial.math.lamar.edu/Classes/CalcII/Series_Special.aspx)

I'm really confused. I failed a test mainly on Series/Sequences, and I need help badly. Thank you.

Answer 1

For the second series, namely $\sum_{n=0}^\infty \frac{1}{n^2+3n+2}$, the general terms of the series approach $0$ as $n\to \infty$.

This does not imply by itself that the series converges. On the other hand, had these general terms not approached zero, then we would conclude that the series diverges.

This series does in fact converge, and we can actually evaluate the series. To do so, we write

$$\begin{align} \sum_{n=0}^\infty \frac{1}{n^2+3n+2}&=\lim_{N\to \infty}\sum_{n=0}^N \frac{1}{n^2+3n+2}\\\\ &=\lim_{N\to \infty}\sum_{n=0}^N \left(\frac{1}{n+1}-\frac{1}{n+2}\right)\\\\ &=\lim_{N\to \infty}\left(1-\frac{1}{N+1}\right)\\\\ &=1 \end{align}$$

However, we need not evaluate the integral to determine whether it is convergent or not. Indeed there are a host of tests, such as the comparison test and the integral test, that can be applied to show that our series of interest is convergent.

Applying the comparison test, we see that $\frac{1}{n^2+3n+2}\le \frac{1}{n^2}$.

Note that $\sum_{n=1}^\infty \frac{1}{n^2}$ converges (See here for a variety of ways to show this). Then, we have

$$0\le \sum_{n=1}^N \frac{1}{n^2+3n+2}\le \sum_{n=1}^N \frac{1}{n^2}\le \sum_{n=1}^\infty \frac{1}{n^2} \tag 1$$

and we see that the sequence of partial sums on the left-hand side of $(1)$ is bounded, and since they are also monotonically increasing, the sequence of partial sums (i.e., the series) converges!

Answer 2

A way to think about this is by use of the vanishing condition:

If $\sum_{i=0}^\infty a_i $ is given to be convergent then the sequence $(a_n)$ is a null sequence.

The contrapositive of this statement is very useful and that's exactly what you used to prove divergence in your first question.

Note, the converse of the vanishing condition is not always true.

For example consider the sequence $a_n= \frac{1}{\ln(n+1)}$. This is a null sequence, but if you look at its graph you'll see it converges so slowly that the series $\sum_{i=0}^\infty\frac{1}{\ln(n+1)} $ is divergent.