What is the sum of this infinite series?

To have a sum, you’d need to know when to stop adding, an end point. An infinite problem, by definition, does not have that. A similar problem might be, what’s 1+2+3+4… forever.

Well, as this progression travels further out, the overall sum will increase less and less, so you will be able to make a statement about the sum approaching a given value as n approaches infinity…

Well, sinch the series is infinite, I presume that the sum is also infinite.

@yankeetooter – It is a bit like Zeno’s Paradox, eh?
http://en.wikipedia.org/wiki/Zeno's_paradoxes#The_dichotomy_paradox

1/n (harmonic series) diverges (i.e. sum goes infinite as number of terms goes to infinity). This happens despite the fact that the individual terms approach zero as n goes to infinity. But there certainly are infinite series with finite sums.

http://www.jimloy.com/algebra/series.htm
1.2020569032 if you are starting from n = 1.

Since this is sum is a p-series with p>1, it does converge. Unfortunately, there is no precise way to arrive at the limit of a p-series sum, so it must be approximated. If you want to approximate it yourself, you can look up methods to do so. I simply utilized Wolfram Alpha to find that the limit of this series is approximately 1.20206.

@laureth To have a sum, you’d need to know when to stop adding, an end point. An infinite problem, by definition, does not have that. Actually, many infinite series do converge to a specific number. Of those, some can be solved precisely, but most must be approximated. Fortunately, they can be approximated to any desired margin of error (except zero error, of course).

@hiphiphopflipflapflop While you are correct about the harmonic series, that is not the series in question. The OP asked about n^(-3), not n^(-1).

@bobbinhood I am well aware of that, but it is a good example of a series of which the sum goes infinite despite the individual terms going to zero. You’ll see your answer checks out with the one I followed up with.

@hiphiphopflipflapflop You edited to add that number. It wasn’t there when I responded. Sorry about that. I still fail to see the relevance of your first answer and your link given that this question is posted in general, and they don’t answer the question that was asked.

The sum of a convergent infinite series is the limit of the partial sums: a0, a0+a1, a0+a1+a2….

@bobbinhood look at the responses by @yankeetooter and @YoBob. Because of them I thought there was some scope for discussion of the properties of sums of infinite series in general. The answer to the OP’s specific question can indeed be found in that link. Sums for 1/n^x series are given for values of x between one and six at about the middle of the page. I edited in the given value for the sake of clarity.

Somewhere between ¼ and ½, which is probably not of much use to you. I did a comparison to the integrals starting at 1 for an upper bound and starting at 2 for a lower bound

The sum 1 + 1 / 8 + 1 / 27 + ... is also known as Apéry’s constant, which is Z(3) where Z = Riemann zeta function. The actual value is about 1.202056903159594285399738161511449990764986292 (already given by @hiphiphopflipflapflop).

Interesting, the reciprocal of this number is the probability that any three positive integers chosen at random are relatively prime (reference: same as above).

Thanks for the education, especially the link from @hiphiphopflipflapflop.

I agree with @bobbinhood – any series of the form 1/(n^x) for x>1 converges.

And as discussed above, it is possible to add an infinite number of terms and get a finite sum – that is what is meant by convergence. If I recall correctly, any series for which the Nth term approaches zero as N approaches infinity converges. There are many other ways to test for convergence, too.

@Mariah, as @hiphiphopflipflapflop pointed out, the harmonic series is a counter-example to your conjecture: “any series for which the Nth term approaches zero as N approaches infinity converges.”

Oof. Of course, thank you. Now I remember what the rule is. Every series for which the Nth term does not approach zero as N approaches infinity definitely diverges. The inverse is not necessarily true though.

Ok, so this one has been bugging me all day.

I know this is probably explained in the link provided by @hiphiphopflipflapflop. But as I have to deal with that work thing I don’t have time to digest much more than the 15 second sound byte of any new info these days. I’m hoping to get the “for dummies” version.

I can wrap my puny brain around the whole concept of a convergence point. But I am having trouble understanding the upper boundary of this point. I understand that as n approaches infinity the corresponding term in the series approaches zero. However, that term will always remain a non-zero value so I don’t quite get how to define the upper boundary.

The value requested by the OP is Avery’s constant – 1 =.202…, since the requested series starts with ½^3.

@YoBob

To understand convergence a bit better, consider the number 1/9, which in decimal form is .1111111… off into infinity. You can think of this as an infinite sum:

.1 + .01 + .001 + .0001 + ... + 1/10^n. This, like the example in the OP, is an example of a p series with p>2, which by definition converge. We also know it converges because we know it equals 1/9.

When you add .01 to .1 you get .11, and with the next term you get .111, and no matter how many terms you add you will never get up to .2, because each term is getting smaller. Each subsequent term is small enough that it only affects the next decimal place (first the tenths then the hundredths then the thousandths, etc.) and you’ll never be able to add enough to affect a previous decimal place. That’s the best way I can think to convey it.

Thank you @Mariah. An excellent explanation!

@PhiNotPi hope you don’t mind me asking: how did this problem crop up for you?

@gasman that’s a very interesting connection to the distribution of the prime numbers among the integers.

@Mariah That is a very good example of when a series converges. On a side note, that is not a p-series. A p-series has an increasing number to a constant power, while your series had a constant number to an increasing power. A p-series with p=10 would look like 1/n^10.

@hiphiphopflipflapflop It really didn’t crop up for me. I create my own math problems to challenge myself. The ones I can’t solve end up on Fluther.

@PhiNotPi Argh, so many mistakes in this thread! You’re right of course, thank you. This material was covered in my first college calculus class and doesn’t seem to have sunk in too well yet.

@Mariah After it sinks in it begins to ooze back out again…
0.111… is a beautifully clear example of an exponential series converging to a finite limit.
0.999… may elicit some unease, though mathematically it’s unambiguously equal to one…