What is the difference between convergence of a sequence and convergence of a series?

You can identify a series with the sequence of its partial sums: $$ S_n = \sum_{k=1}^n a_k. $$ So everything you know about sequences can be applied to series, and vice-versa.

However dealing with series is usually more difficult because, in general, it can be very difficult to find the limit. This is due to the indirect definition of partial sums... So to deal with series and to prove their convergence, one should use methods which do not require the limit (i.e. the sum) of the series to be known in advance.

Series can be baffling things. The trouble is that the convergence of the terms tells you nothing about the convergence of the series. We know that $\lim_{n\to\infty}a_n=0$ is necessary for a series $\sum_{k=0}^\infty a_n$ to converge, but it is not sufficient.

For example consider the sequences $a_n=1/n$, $b_n=(-1)^n/n$, $c_n=1/n^2$, $d_n=1/n^3$, $e_n=1/n^5$. All of these sequences converge to zero. But:

• $\sum_{k=1}^\infty a_n$ diverges (harmonic series)
• $\sum_{k=1}^\infty b_n$ converges (alternating series)
• $\sum_{k=1}^\infty c_n$ converges to $\pi^2/6$
• $\sum_{k=1}^\infty d_n$ converges, to an irrational number (Apéry's theorem)
• $\sum_{k=1}^\infty e_n$ converges, but its not known if the result is rational or irrational.

There is, as far as I know, no complete decision method for convergence of a series. You try a sequence of tests, and each one will return a result of converge, diverge, undecided. If the latter, you try another test.

There are certain families of series, such as geometric series and $p$-series, for which convergence or divergence is trivial ($\sum_{k=0}^\infty r^k$ converges if and only if $|r|<1$), but in general, given a new series which doesn't fit into a known class, you're on your own.

As far as I know, this problem is unsolved: is there a sequence of rational numbers $a_n$ for which $\lim_{n\to\infty}a_{n+1}/a_n=0$ and $\sum_{n=0}^\infty a_n=\pi$? (Note that if we replace $\pi$ by $e$ then the result is trivial, as we can put $a_n=1/n!$.)

As you see, series are tricky.

If we are talking about sequences and series of real or complex numbers, or of vectors in a real (or complex) normed vector space, then convergence of sequences and series are equivalent concepts.

Convergence of a series $\sum_{n=1}^\infty a_n$ is simply the convergence of the sequence of partial sums $S_N = \sum_{n=1}^N a_n$.

Convergence of a sequence $(a_n)$ of numbers is equivalent to the convergence of the series $a_1 + \sum_{n=1}^\infty (a_{n+1} - a_n)$; note the $N$-th partial sum is $$ a_1 + \sum_{n=1}^N (a_{n+1} - a_n) = a_1 + \sum_{n=2}^{N+1} a_n - \sum_{n=1}^N a_n = a_1 + a_N - a_1 = a_N. $$

However, in the more general setting of topology the notion of a sequence is more general than the notion of a series, because in topological spaces without a vector space structure the notion of a series makes no sense (how will you define a series if you can't add things?). Therefore it may be convenient to think of sequences as the more natural idea, and series as a special case of sequences.