Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
3  
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
2  
@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.

share|improve this answer

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 $\beta>0$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.