Why does the series 1 n converge

This will always be true for convergent series and leads to the following theorem. Then the partial sums are,. Be careful to not misuse this theorem!

This theorem gives us a requirement for convergence but not a guarantee of convergence. In other words, the converse is NOT true. Consider the following two series. The first series diverges. Again, as noted above, all this theorem does is give us a requirement for a series to converge. In order for a series to converge the series terms must go to zero in the limit.

If the series terms do not go to zero in the limit then there is no way the series can converge since this would violate the theorem. Again, do NOT misuse this test.

If the series terms do happen to go to zero the series may or may not converge! Again, recall the following two series,. There is just no way to guarantee this so be careful! The divergence test is the first test of many tests that we will be looking at over the course of the next several sections. You will need to keep track of all these tests, the conditions under which they can be used and their conclusions all in one place so you can quickly refer back to them as you need to.

Furthermore, these series will have the following sums or values. At this point just remember that a sum of convergent series is convergent and multiplying a convergent series by a number will not change its convergence. We need to be a little careful with these facts when it comes to divergent series. Now, since the main topic of this section is the convergence of a series we should mention a stronger type of convergence.

Absolute convergence is stronger than convergence in the sense that a series that is absolutely convergent will also be convergent, but a series that is convergent may or may not be absolutely convergent. When we finally have the tools in hand to discuss this topic in more detail we will revisit it. The idea is mentioned here only because we were already discussing convergence in this section and it ties into the last topic that we want to discuss in this section. First, we need to introduce the idea of a rearrangement.

A rearrangement of a series is exactly what it might sound like, it is the same series with the terms rearranged into a different order. The issue we need to discuss here is that for some series each of these arrangements of terms can have different values despite the fact that they are using exactly the same terms.

The values however are definitely different despite the fact that the terms are the same. Here is a nice set of facts that govern this idea of when a rearrangement will lead to a different value of a series.

This is here just to make sure that you understand that we have to be very careful in thinking of an infinite series as an infinite sum. There are times when we can i. Mathematicians say that the series "diverges to infinity".

I think the integral test gives the most intuitive explanation. The proof I learned, in Rosenlicht's Introduction to Analysis , published by Dover, is essentially a variant of the most popular answers above. Beginning with the equation.

Thus the harmonic series diverges. I think the greedy algorithm for constructing Egyptian fractions doesn't work here, as you already need to know that the harmonic series diverges in order to prove that the greedy algorithm works for arbitrarily large rationals This is based on the same idea as several other answers, but the presentation, I think, is sufficiently different to make it worth adding. Sign up to join this community. The best answers are voted up and rise to the top. Stack Overflow for Teams — Collaborate and share knowledge with a private group.

Create a free Team What is Teams? Learn more. Ask Question. Asked 11 years, 3 months ago. Active 1 year ago. Viewed 21k times. Does this give a hint? This proof is easily comprehensible if you know the dominated convergence theorem, but that theorem is not the most comprehensible. Show 1 more comment. Active Oldest Votes. This proof is often attributed to Nicole Oresme. Tunk-Fey AgCl AgCl 5, 7 7 gold badges 37 37 silver badges 37 37 bronze badges.

Wikipedia has a nice display of this proof [ en. Show 4 more comments. EDIT It seems the original link is broken, due to the author moving to his own site. Kifowit and Terra A. Add a comment. Ian Mateus Ian Mateus 6, 3 3 gold badges 29 29 silver badges 54 54 bronze badges. And possibly pedagogically best; I don't know.

Dominic Michaelis Khosrotash Khosrotash What's the reasoning their Noah Snyder Noah Snyder 9, 3 3 gold badges 36 36 silver badges 58 58 bronze badges. I was hoping to avoid an answer that involved integration, so I also prefer AgCl's answer. The proof is complete. Teoc 8, 3 3 gold badges 22 22 silver badges 55 55 bronze badges. Ahaan S.

Rungta Ahaan S. Rungta 7, 5 5 gold badges 25 25 silver badges 63 63 bronze badges. Rasmus Erlemann Rasmus Erlemann 2, 2 2 gold badges 16 16 silver badges 30 30 bronze badges.

Then this contradiction shows that the series diverges. Jose Antonio Jose Antonio 6, 2 2 gold badges 23 23 silver badges 37 37 bronze badges. Squirtle Squirtle 6, 27 27 silver badges 55 55 bronze badges. Ryan Reich Ryan Reich 6, 19 19 silver badges 30 30 bronze badges. The reason that it does not work is because the number of terms is infinite.

All you've proven is that the second series approaches the final value faster than the first one. But whether this value is ultimately finite or not, you have not shown. The student's question was