So all I could get from my teachers thick accent in class today is that:

A sequence is finite and converges when bounded by x?


A series is infinite and diverges because no matter how small the function gets, it will never reach zero?

I'm sorry. I'm really having a hard time understanding the concept. For example. This picture... ! $$\begin{align} s_1 &= a_1\\ s_2 &= a_1+a_2\\ s_3 &= a_1+a_2+a_3\\ s_4 &= a_1+a_2+a_3+a_4\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\vdots\\ s_n &=a_1+a_2+a_3+a_4+\cdots+a_n=\sum_{i=1}^na_i \end{align}$$

I don't the last part involving the sigma. How is that to the statments before it.

  • 1
    $\begingroup$ A sequence is a set of numbers that (typically) have a function that relates them. A series is a sum over a sequence. Both sequences and series can be convergent or divergent depending on the function describing them. If a series converges, then so does the sequence it is based on; if a sequence diverges, then so does a series based on it. The capital sigma $\Sigma$ is the usual notation to say "$\sum_{i=1}^na_i$ is the sum from $1$ to $n$ of the numbers $a_i$, i.e., $a_1+a_2+a_3+\cdots+a_{n-1}+a_n$." $\endgroup$
    – abiessu
    Oct 24, 2013 at 21:48
  • $\begingroup$ Watch these videos, and see if it helps: khanacademy.org/math/calculus/sequences_series_approx_calc/… $\endgroup$
    – apnorton
    Oct 24, 2013 at 21:53
  • $\begingroup$ The last equality is true by definition of the Sigma-notation ("$\Sigma$"). But your two statements about sequences and series sound like nonsense to me, out of context anyway. Would it help to sit closer to the teacher? If not, I would suggest skipping class and reading a book instead. $\endgroup$ Oct 24, 2013 at 21:56
  • $\begingroup$ "So all I could get from my teachers thick accent in class today is..." Looking at what you wrote, I guess that the teacher's accent is the least of your problems in this class. $\endgroup$
    – fedja
    Oct 24, 2013 at 23:00
  • $\begingroup$ I still don't understand the A(n-1) + An thing. I just don't understnad the logic behind that notation. $\endgroup$
    – Josh
    Oct 25, 2013 at 19:21

1 Answer 1


In textbooks, the convergence of a sequences is often discussed before series.

If a sequence is monotonous, say $a_{n+1} \geq a_n$, and bounded from above, yes, it will converge. This is proven by showing it has an upper bound (reformulation of "bounded from above"), and using the fact that every bounded set has a smallest upper bound (called supremum). This smallest upper bound of the sequence is its limit.

Next, by considering the partial sums of a series, you will notice that the partial sums form a sequence. If this sequence of partial sums converges, the series is said to converge.

Your textbook seems to be confusing for you for not repeating the partial sum notation in the lines you quote. Just consider that writing the partial sum with a sum-sign is the shorthand for the many terms of the sum added together. The $s_n$ form the sequence of partial sums of the $a_n$.

If a sequence of partial sums $s_n$ converges, it follows that the sequence of its terms $a_n$ converge to 0. Proof. For all n>N, $|s_n - s_{n+1}|$ will be arbitrarily small, hence $|a_{n+1}|$ will be arbitrarily small, i.e., the sequence of $a_n$ converges to 0.

Caution. The reverse is not true. If the sequence $(a_n)_{n \geq 1}$ converges to 0, then it does NOT follow that the sequence of its partial sums $(s_n)_{n \geq 1}$ converges.

Example. The sequence $(a_n)_n = (1/n)_n$ converges to 0, however the sequence of the partial sums $s_N = \sum_{n=1}^N a_n$ is unbounded (hence, the series diverges). Note that $\sum_{n=2^{k}+1}^{2^{k+1}} 1/n > 1/2$ for all k.


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