Talk:Convergent series

Mistake?

The statement of the Cauchy criterion for convergence is incorrect, but I am not sure how it should be changed. David Radcliffe 19:35, 30 November 2005 (UTC)

I think it is correct just fine. :) It says that the sequence of partial sums is a Cauchy sequence, which is equivalent to the sequence being convergent. Oleg Alexandrov (talk) 00:20, 1 December 2005 (UTC)

I'm sorry, but it does not currently state that the partial sums form a Cauchy sequence. I will post a correction. David Radcliffe 02:50, 1 December 2005 (UTC)

You are right, it was close to Cauchy sequences, but not quite. The harmonic series would have satisfied that criterion which was wrong. I also posted a formulation after you fixed that part. Oleg Alexandrov (talk) 03:10, 1 December 2005 (UTC)

Thanks. I like your formulation using double limits. David Radcliffe 03:39, 1 December 2005 (UTC)

More convergence tests

Unde the convergence tests section, the article could also mention the n-th term test, the p-series, the geometric series, and most importantly, the power series. If anyone willing to add these tests, please do so. Otherwise I'll add them in later on.--Rejnal 04:33, 15 March 2006 (UTC)

Yes, please add them. If you feel you have a lot of detail to put in, you could also create new articles about those topics. Cheers, Oleg Alexandrov (talk) 04:37, 15 March 2006 (UTC)

Can't the limit comparison test also establish divergence iff b sub n diverges. The test is inconclusive if the limit does not exist. I think it should be mentioned. Good luck! Cronholm144 23:35, 6 May 2007 (UTC)

Real numbers, complex numbers? A few suggestions I do not have time to work on myself

I just skimmed the article and did not see any mention of real numbers or complex numbers or anything. So, do all of the tests work for complex numbers? I am pretty sure not all work for complex numbers, but many do if written appropriately. So, here are a few suggestions. The first 3 are basically the same idea, the 4th is different.

1. Mention that we are talking about complex and real numbers here.

2. For each test, or each part of the article, be specific about which of the two types of numbers it works for.

3. Make things more general. The ratio test, for example, as stated only works for series which are all positive real numbers. But, in fact it works for series with positive and negative terms and even complex series. It should be stated more generally, I think, so that it is more useful.

4. An even more general section could be added which talks about convergent series in a general topological space. I believe this article is probably mostly accessed by high school and college students who do not know a whole lot about that, so it might be best to put this new section near the end so as to not scare off people who are just trying to learn about basic series for real numbers. But, it seems fitting for it to at least be mentioned. Maybe there is another article that could be referenced at least. Or, don't add this section and entitle this article "Convergent series of real and complex numbers".

StatisticsMan (talk) 17:33, 28 February 2009 (UTC)

The discussion here is all real. No section can be added about convergent sequences in topological spaces because in general there is no such thing. Sequences can converge (consider neighbourhoods around points), but in general there is no operation of summation, which is clearly needed for series. Clearing up the complex cases a bit is a good idea though.— Kan8eDie (talk) 14:04, 2 March 2009 (UTC)

equality

I think the equality signs should be changed to -> or < or > as the case may be. Obviously the series never takes the value so it can never be an equality. So the questions "find the sum of the series.." can be treated as flawed in most cases can it not? —Preceding unsigned comment added by 192.18.192.76 (talk) 07:21, 20 July 2009 (UTC)

• The symbols are correct. The value of an infinite series is defined as its limit, so that $(1+{\tfrac {1}{2}}+\cdots )$ is shorthand notation for $\lim _{n\rightarrow \infty }\sum _{k=0}^{n}2^{-k}$, and a limit is just a plain old real number with the property that eventually all partial sums stay arbitrarily close to it. So, there is no equality for the partial sums (you are right in saying that for any n, $\sum _{k=0}^{n}2^{-k}\neq \sum _{k=0}^{\infty }2^{-k}$), but there is equality for the whole, limiting, sum (the equality $\sum _{k=0}^{\infty }2^{-k}=2$ really does hold). The question "find the sum of the series" is thus not a flawed question, as it just means "find the value of the limit of the sequence of partial sums", which is a good question with a real, well-defined, answer if there is convergence.— Kan8eDie (talk) 15:24, 20 July 2009 (UTC)

Thank you for your comments, if the definition of the "limit" is that there exists an e>0 such that |xn-L|<e , hence L being the limit of xn, then the same logic should apply for series convergence too. The series never will attain that value, do you agree or not? —Preceding unsigned comment added by Alokdube (talk • contribs) 06:36, 21 July 2009 (UTC)

More general notions of convergence?!

Why in the world does Convergence (mathematics) redirect here?! Series convergence is just a special case of sequence convergence, which is a special case of convergence of nets, which can further be generalized using filters. In other words, the notion of convergence is much more general than convergence of series, so there really should be a separate page called Convergence (mathematics) that discusses the general concept. Actually, it might make more sense for Convergence (mathematics) to just redirect to Limit (mathematics) (instead of Convergent series) since the concepts of "convergence" and "limit" are essentially the same, but I don't know how to do redirects. 216.160.76.121 (talk) 03:19, 30 August 2009 (UTC)

I agree, I redirected to limit (mathematics)--Work permit (talk) 05:18, 30 August 2009 (UTC)

I read about the relationship between the zeta function and the eta function. The eta function has a domain of x > 0, whilst the zeta function's domain is only greater than 1. However, it is possible to express them as ζ(s) = η(s) / (1-(1/2^s-1)), implying that the zeta function has values for all positive arguments. Why is this, and should this sort of thing be included in the article? My 2 Cents' Worth (talk) 14:12, 30 July 2010 (UTC)

The Riemann zeta function is defined by analytic continuation of the series Σ n^−s. Bo Jacoby (talk) 18:58, 30 July 2010 (UTC).

Algebraic example

Might it help if there was a more general, algebraic example? I have recently added

• And, when 1 < x and 1 < y, we have a more general convergent series, using algebra:

  ${x \over y^{0}}+{x \over y^{1}}+{x \over y^{2}}+{x \over y^{3}}+{x \over y^{4}}+{x \over y^{5}}+\cdots =x(1{1 \over y-1}).$

to the bottom of the Examples of convergent and divergent series section, but it was quickly reverted by GraemeMcRae (talk | contribs) with the edit summary 'reverting change by Robo37. Not only is it wrong (x need not be greater than 1, and the sum should have been xy/(y-1)) but it's unsourced and nothing more than a general geometric series.'

First of all, it was not wrong at all, okay so maybe x doesn't have to be greater than 1, but that doesn't make it wrong. I was in fact planning to remove '1 < x and', but GraemeMcRae removed everything before I had the chance to. And as for the formula itself, well we have

• ${1 \over 2^{0}}+{1 \over 2^{1}}+{1 \over 2^{2}}+{1 \over 2^{3}}+{1 \over 2^{4}}+{1 \over 2^{5}}+\cdots =1_{2}+0.1_{2}+0.01_{2}+0.001_{2}+0.0001_{2}+0.00001_{2}+\cdots =$

  $1.11111..._{2}=1{1 \over 1}=2,$

• ${1 \over 3^{0}}+{1 \over 3^{1}}+{1 \over 3^{2}}+{1 \over 3^{3}}+{1 \over 3^{4}}+{1 \over 3^{5}}+\cdots =1_{3}+0.1_{3}+0.01_{3}+0.001_{3}+0.0001_{3}+0.00001_{3}+\cdots =$

  $1.11111..._{3}=1{1 \over 2},$

• ${1 \over 4^{0}}+{1 \over 4^{1}}+{1 \over 4^{2}}+{1 \over 4^{3}}+{1 \over 4^{4}}+{1 \over 4^{5}}+\cdots =1_{4}+0.1_{4}+0.01_{4}+0.001_{4}+0.0001_{4}+0.00001_{4}+\cdots =$

  $1.11111..._{4}=1{1 \over 3},$

• ${1 \over 5^{0}}+{1 \over 5^{1}}+{1 \over 5^{2}}+{1 \over 5^{3}}+{1 \over 5^{4}}+{1 \over 5^{5}}+\cdots =1_{5}+0.1_{5}+0.01_{5}+0.001_{5}+0.0001_{5}+0.00001_{5}+\cdots =$

  $1.11111..._{5}=1{1 \over 4},$

and so on.

The effect that x has on this is clear, for example when x = 2 each fraction is doubled, so the answer must also be doubled. We have:

• ${2 \over 5^{0}}+{2 \over 5^{1}}+{2 \over 5^{2}}+{2 \over 5^{3}}+{2 \over 5^{4}}+{2 \over 5^{5}}+\cdots =2_{5}+0.2_{5}+0.02_{5}+0.002_{5}+0.0002_{5}+0.00002_{5}+\cdots =$

  $2.22222..._{5}=2{2 \over 4}=2{1 \over 2}.$

I find this a lot more notable than the other examples; it shows the reader what the sum of reciprocals of every group of powers is, not just those belonging to the powers of two. And it's essentially the same thing as saying that 0.11111... is equal to 1/9 so I doubt it needs to be sourced, and if a source is needed that badly, it'll be easy to find.

Also note that

• ${9 \over 10^{1}}+{9 \over 10^{2}}+{9 \over 10^{3}}+{9 \over 10^{4}}+{9 \over 10^{5}}+\cdots =0.9+0.09+0.009+0.0009+0.00009+\cdots =0.99999...=1.$

(see 0.999...)

Robo37 (talk) 13:09, 10 November 2009 (UTC)

And if it needs to be simplfied you can also have

• ${1 \over y^{0}}+{1 \over y^{1}}+{1 \over y^{2}}+{1 \over y^{3}}+{1 \over y^{4}}+{1 \over y^{5}}+\cdots =1{1 \over y-1}$

or

• ${1 \over y^{1}}+{1 \over y^{2}}+{1 \over y^{3}}+{1 \over y^{4}}+{1 \over y^{5}}+\cdots ={1 \over y-1}.$

Robo37 (talk) 13:21, 10 November 2009 (UTC)

Sorry, I could have probably worded that better. I fail at this kinda thing. Well anyway... does anyone object to me putting

• ${x \over y^{0}}+{x \over y^{1}}+{x \over y^{2}}+{x \over y^{3}}+{x \over y^{4}}+{x \over y^{5}}+\cdots =x(1{1 \over y-1}),$

• ${1 \over y^{0}}+{1 \over y^{1}}+{1 \over y^{2}}+{1 \over y^{3}}+{1 \over y^{4}}+{1 \over y^{5}}+\cdots =1{1 \over y-1}$

or

• ${1 \over y^{1}}+{1 \over y^{2}}+{1 \over y^{3}}+{1 \over y^{4}}+{1 \over y^{5}}+\cdots ={1 \over y-1}.$

somewhere in this article? Robo37 (talk) 09:29, 17 November 2009 (UTC)

I object to your use of $1{1 \over y-1}$ because, while you mean to add 1 to ${\frac {1}{y-1}}$, the normal convention would be to multiply them! $1+{1 \over y-1}$ would be a better way to write this, if it is used in the article.

---Glenn L (talk) 14:23, 17 November 2009 (UTC)

hmmm... it might be better to just leave the ${1 \over y^{0}}$ bit (the 1) out so it looks better. We then have

• ${x \over y^{1}}+{x \over y^{2}}+{x \over y^{3}}+{x \over y^{4}}+{x \over y^{5}}+\cdots ={x \over y-1}$

or

• ${1 \over y^{1}}+{1 \over y^{2}}+{1 \over y^{3}}+{1 \over y^{4}}+{1 \over y^{5}}+\cdots ={1 \over y-1}.$

Robo37 (talk) 15:16, 17 November 2009 (UTC)

Or, you could rewrite the other two as:

${x \over y^{0}}+{x \over y^{1}}+{x \over y^{2}}+{x \over y^{3}}+{x \over y^{4}}+{x \over y^{5}}+\cdots ={xy \over y-1}$

and

${1 \over y^{0}}+{1 \over y^{1}}+{1 \over y^{2}}+{1 \over y^{3}}+{1 \over y^{4}}+{1 \over y^{5}}+\cdots ={y \over y-1}.$

Six convergent series needing solving.

1. What is the sum of the reciprocals of the super squares?

• ${1 \over 1^{1}}+{1 \over 2^{2}}+{1 \over 3^{3}}+{1 \over 4^{4}}+{1 \over 5^{5}}+{1 \over 6^{6}}+\cdots =?.$

2. What is the sum of the reciprocals of the super powers of two?

• ${1 \over 2}+{1 \over 2^{2}}+{1 \over 2^{2^{2}}}+{1 \over 2^{2^{2^{2}}}}+{1 \over 2^{2^{2^{2^{2}}}}}+{1 \over 2^{2^{2^{2^{2^{2}}}}}}+\cdots =?.$

3. What is the sum of the reciprocals of the Partion numbers?

• ${1 \over 1}+{1 \over 2}+{1 \over 3}+{1 \over 5}+{1 \over 7}+{1 \over 11}+\cdots =?.$

4. What is the sum of the reciprocals of the Ulam numbers?

• ${1 \over 1}+{1 \over 2}+{1 \over 3}+{1 \over 4}+{1 \over 6}+{1 \over 8}+\cdots =?.$

5. What is the sum of the reciprocals of the Superfactorials?

• ${1 \over 1!}+{1 \over 1!\times 2!}+{1 \over 1!\times 2!\times 3!}$+${1 \over 1!\times 2!\times 3!\times 4!}+{1 \over 1!\times 2!\times 3!\times 4!\times 5!}+{1 \over 1!\times 2!\times 3!\times 4!\times 5!\times 6!}+\cdots =?.$

6. What is the sum of the reciprocals of the triangular Factorial numbers?

• ${1 \over 1!}+{1 \over 1!+2!}+{1 \over 1!+2!+3!}+{1 \over 1!+2!+3!+4!}+{1 \over 1!+2!+3!+4!+5!}+{1 \over 1!+2!+3!+4!+5!+6!}+\cdots =?.$

I'm looking for closed forms. Robo37 (talk) 10:51, 15 August 2011 (UTC)

Reciprocals of primes

How did Euler prove the sum of the reciprocals of the primes diverges? I know he created the Riemann zeta function for it but how did he do this? — Preceding unsigned comment added by 86.185.99.78 (talk) 21:18, 11 December 2012 (UTC)

Conditionally convergent

The following sentence is misleading: The power series of the logarithm is conditionally convergent. The power series of log(1+z) around 0 is absolutely convergent for |z|<1, divergent for |z|>1 and for |z|=1 various things can happen (the radius of convergence is 1).

Sprocedato (talk) 11:26, 21 January 2013 (UTC)