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Assume I have a complex sequence $(a_k)_{k\ge1}$ with $\lim_{k \to\infty} a_k = 0$ and let $\sum_{k=1}^\infty a_k$ be the corresponding series. It is intuitively obvious the this series converges if the $a_k$ tend to zero fast enough. Can this be made more precise? Can I assign to a sequence a "speed" of convergence, and then determine whether this is adequate for convergence of the series? thanks

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Did you mean $\lim_{k\rightarrow +\infty}a_k=0$ then? –  1015 Feb 3 '13 at 21:12
The most precise way, as far as I know, is to say it as a definition: a nullsequence $a_n$ has the 'finitely summable' convergence speed iff $\sum a_n$ converges. –  Berci Feb 3 '13 at 21:13
@Amir: The answer is no, as there is no "smallest" diverging series. However, you can often decide convergence by comparing against other series, e.g. $\sum \frac 1 n$ diverges, but $\sum \frac 1 {n^s}$ for $s>1$ converges, as does $\sum q^n$ for $|q|<1$. –  Dario Feb 3 '13 at 21:18

2 Answers 2

Since you are asking about complex sequences and not positive sequences, the idea of conditional convergence arises. $$ \sum_{k=2}^\infty\frac{(-1)^k}{\log(k)}\tag{1} $$ converges, but $$ \sum_{k=2}^\infty\frac1{k\log(k)}\tag{2} $$ diverges, even though the terms in $(2)$ tend to $0$ more quickly.

If we focus on positive sequences, even if we define the rate of convergence to $0$ as $$ s_n=\sup_{k\ge n}a_k\tag{3} $$ Then for the sequence $$ a_k=\left\{\begin{array}{} \frac1{k^2}&\text{if }k\text{ is not a power of }2\\ \frac2{k}&\text{if }k=2^j \end{array}\right.\tag{4} $$ we have $$ \sum_{k=1}^\infty a_k=\frac{\pi^2}{6}+\frac83\lt\infty\tag{5} $$ yet $\frac1n\lt s_n\le\frac2n$, which is greater than that for the harmonic series, which diverges.

For monotonically decreasing, positive sequences, we do have the comparison test, which defines how fast something goes to $0$ by whether the series converges.

As Dario comments, there is no smallest diverging series or largest converging series.

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One cannot determine convergence of the series by looking at only a few terms. In fact, one can always throw out any finite number of terms, and convergence is not affected. So any criteria for fast enough decay of $a_k$ must involve all but finitely many terms.

By comparing with well-known converging series, one can give sufficient criteria for convergence.

It is easy to see, that if $a_k \geq C/k$ for some constant $C$, then the series diverges. On the other hand, we know that $1/k^\alpha$ with $\alpha>1$ always converges.

The little-oh notation is very useful here: We write $a_k = o(b_k)$ if an only if $$ \lim_k \frac{|a_k|}{|b_k|} = 0, $$ saying that "$a_k$ is of order $b_k$". For example, any absolutely convergent series must satisfy $a_k = o(1/k)$. All sequences with $a_k = o(1/k^\alpha)$ with $\alpha > 1$ converges. The $\alpha = 1$ case is a limiting case.

The number $\alpha$ can be used to measure how well a partial sum approximates the total sum, i.e., a measure of the "speed of convergence". The larger $\alpha$ is, the better a partial sum will approximate the total.

Note, however, that there are sequences which are $o(1/n)$ but not $o(1/n^\alpha)$ for any $\alpha>1$, so the proposed classification is not exhaustive. Moreover, there are series with terms that are $o(1/k^\alpha)$ for any $\alpha>1$, e.g., $o(\exp(-\beta k))$ for some $\beta>0$.

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