The question is why the series diverges. If you can reconcile how this works for ln(x) you can understand the harmonic series. Since 1> 1/2, 1 + 1/2 > 1/2 + 1/2 = 1. You have to look at ALL terms to determine converge/divergence. A quitté FuturaSciences. Il est actuellement, Futura-Sciences : les forums de la science, http://fr.wikibooks.org/wiki/Analyse:S%C3%A9ries. So, pick some N and some sum up to that 1/N. So the previous result is perfectly coherent with H_n ∈ ln n + γ + O(1/n) . There is 2n terms in this sums, and its smallest term is 2-(n+1). The second is 1/2, also leave that. We're never going to be adding 0 to it. So, pick some N and some sum up to that 1/N. If we subtract some amount from those terms (calculating the exact amount doesn’t matter), we can change those terms to all be 1/8. Comment définir ou déterminer cette fonction? So this is just circular reasoning. In fact, we can get a measure of this difference and ln(N) as N->infinity and this is the Euler-Mascheroni Constant. You can think of the "last" term of the sequence being epsilon, arbitrarily close to 0 but still bigger than 0 and thus the sum will always be getting bigger. So the first group is strictly larger than 1. Not only is it not right, it's not even wrong! The first term is 1, so leave that. Thus, the difference D_n is greater than 1/2. The amount we subtracted doesn’t matter because even though we subtracted from the series, it is still divergent, so it will tend towards infinity if we include the amount or not. The next four terms are 1/5, 1/6, 1/7, and 1/8. Press J to jump to the feed. Therefore, it takes us more and more term to add more than 1/2 to the result of the series, but we reach this point eventually. Doesn't the value that you have to substract from your terms to make the groups have equal values get bigger and bigger for every 1/2 that you add? That's the oldest proof we have. This kind of thing is why I clicked on this thread. Bon Plan Prixtel : le forfait Giga Série 50 Go à 12,99 €/mois, Forfait Série Free : bon plan de 70 Go proposé à 10,99 €/mois, FIC 2020 : comment hacker une voiture de série en deux leçons, Le Pipistrel Velis Electro devient le premier avion 100 % électrique de série, Par mattlastar dans le forum Mathématiques du supérieur, Par MiMoiMolette dans le forum Mathématiques du supérieur, Par prgasp77 dans le forum Mathématiques du supérieur, Par carl159 dans le forum Mathématiques du supérieur, Fuseau horaire GMT +1. Then simplify the terms, so the two 1/4’s become 1/2 and the four 1/8’s become 1/2 and so on. But that's just it. As x goes to infinity, ln(x) gets arbitrarily large at ever slower rates. So, (1 + 1/2) + (1/3 + 1/4) + (1/5 + 1/6) + ... > 1 + 1/2 + 1/3 + ... and therefore, S>S, which is impossible. Which means that no matter what value of N you've summed the series up to (even 1/graham's-number) you can always, always, always add more than 1/2 to the sum. we can get a measure of this sum and ln(N) as N->infinity and this is the Euler-Mascheroni Constant. Press question mark to learn the rest of the keyboard shortcuts. The third and fourth is 1/3 and 1/4, but if we subtract a little from that we can change the terms to 1/4 and 1/4. Shouldn't the series converge if we're eventually just going to be adding 0 to it? Then we are left with 1+1/2+1/2+1/2... which is clearly a divergent series. And to make the link with the logarithm function clearer, one can notice that ln 2 > 0.5. Merci de ne PAS me contacter par MP. Série convergente -> série abst convergente, cette solution de serie/parallele/serie fonctionne-t-elle. Or am I misunderstanding the terminology? If you have a function f(x), then the integral of f(x) for x=1 to N will be approximately equal to the sum f(1)+f(2)+...+f(N) (as long as f(x) isn't a terrible function). Each of those terms will be at least as large as 1/2N, and there are N of them, so the sum is at least N/2N, or 1/2. Continue doing this infinitely. Since this integral is infinity, so must be the sum. Likewise, the second group is larger than 1/2, the third group is larger than 1/3, and so on. That's just the definition of what it means for a series to diverge to infinity. You haven't actually explained why it's always possible to choose enough terms to add another +1 to the sum. Ask a science question, get a science answer. Shouldn't the series converge if we're eventually just going to be adding 0 to it? A good illustration is to look at the partial series H_n = 1 + 1/2 … + 1/n and then consider the difference between D_n = H_(2n) - H_(2n+1) = 1/2n + … 1/2n+1 . Very cool, I was not disappointed. Did you mean "a measure of the difference between this sum and ln(N)"? Something a friend told me is a cool trick to think about it. The terms ln(n+1) - ln(n) are all positive go to zero too, yet when you add them up you get a diverging series. And you can do that an infinite number of times, regardless of N. I'd never seen this explanation before, but I really like it! Compare to your knowledge of ln(x). Under what conditions are we allowed to change the grouping of elements of a series ? The short answer is that the series does not decrease fast enough. This means that, for f(x)=1/x, that. There is also a nice visual representation of all this. I get that the integral of 1/x is ln(x) and ln(x) increases with x to infinity, but 1/n also approaches 0 as n approaches infinity. If the series converged then for any given sum up through 1/N the remaining sum beyond that should be getting smaller and smaller, if not then it won't converge. In other words, the partial series grows slowly since multiply by 2 the number of summed terms approximately adds a constant term; but it does grow. New comments cannot be posted and votes cannot be cast. But your answer is essentially "because we can always add enough terms to add at least +1 to the current sum". If the series converged then for any given sum up through 1/N the remaining sum beyond that should be getting smaller and smaller, if not then it won't converge. Indeed, the whole series can be written as the sum of its difference: H_(2n) = D_0 + … + D_(n-1), so for any number x, we know that H_(22 * x) > x . Now, add the sum from 1/N up through 1/2N. Now, add the sum from 1/N up through 1/2N. So S can't actually exist. Oresme came up with it in 1350. Re : série 1/n diverge A ne pas confondre la limite de 1/n en l'infini qui converge, et la somme de terme, certe plus petit les uns après les autres, … If you only looked at the first trillion terms than it pretty much converges to some number because every subsequent addition is negligible, but that's just it. But you can easily fit the graph of 1/(x+1) underneath these blocks, see this. The fractions approach 0, but the fractions are always going to be bigger than 0. If this were the case, the value that you would have to substract from your terms would approach the 1/2 that you are adding to your series, which would bring you back to the original problem where the values that you're actually adding in your series approaches zero.