14
$\begingroup$

Just thinking about it in terms of logic, shouldn't the series of a sequence whose limit as $n$ approaches infinity is 0 converge?

I know that the $n$th term test for divergence says that if a series is convergent, then the limit of its sequence is 0 and I also know there are some sequences for which it has been "proven" that their series does not converge even though the sequence converges to 0, but I just don't believe these tests. If we stretch $n$ out to infinity and the terms are approaching 0, then how is it possible for the sum of the terms to "overflow" and diverge if the terms are becoming negligibly small?

$\endgroup$
  • 54
    $\begingroup$ "but I just don't believe these tests.". Then you should start writing proofs yourself. $\endgroup$ – Stefan Perko Oct 19 '16 at 12:51
  • 22
    $\begingroup$ I wonder what you mean by "I just don't believe these tests". For example, if there is a statement that $\sum_{n=1}^{\infty} 1/n$ does not converge, and if you read a proof for that: what exactly is it that you don't believe? $\endgroup$ – Andreas Oct 19 '16 at 12:52
  • 16
    $\begingroup$ It's quite hard to convince someone who will throw away rigorous proof just because it doesn't "click" for him. $\endgroup$ – Hans Janssen Oct 19 '16 at 14:12
  • 19
    $\begingroup$ I think the key issue here is in your comment that "It seems that it should converge, yet it doesn't." Even though the answers by Henning Makholm and 5xum seem to have solved the present problem for you, you're likely to encounter other situations where your intuition of what should happen disagrees with what a proof shows actually does happen. In any such situation, you should study the proof carefully in order to correct your intuition. Knowing that your intuition was wrong is a good thing, but improving your intuition is even better. $\endgroup$ – Andreas Blass Oct 19 '16 at 17:40
  • 14
    $\begingroup$ I agree with Andreas' comment. The key phrase in your question is the very first words: "just thinking about it in terms of logic" is what you say, but then you go on to say that you wish to reject the rigorous logical conclusions on the basis of an intuition about how things should work. There are many, many counter-to-your-intuition conclusions in mathematics -- particularly in non-finite mathematics! -- and it takes a considerable amount of training to develop an intuition for how it really works. Keep at it! $\endgroup$ – Eric Lippert Oct 19 '16 at 18:23
123
$\begingroup$

A very easy counterexample would be $$ 1, \underbrace{\frac12, \frac12}_{2\text{ halves}}, \underbrace{\frac13, \frac13, \frac13}_{3\text{ thirds}}, \underbrace{\frac14, \frac14, \frac14, \frac14}_{4\text{ fourths}}, \underbrace{\frac15, \frac15, \frac15, \frac15, \frac15}_{5\text{ fifths}}, \ldots $$ This sequence clearly converges to $0$, but if you try to sum it, it should be obvious that it has partial sums as large as you'd like them to be -- so the series diverges.

Try whichever argument you have in mind for believing that the series should converge, and attempt to figure out why it doesn't work for this one.

$\endgroup$
  • 4
    $\begingroup$ Isn't it (simillar to) part of a proof that $\sum_n^\infty\frac1n$ is divergent? $\endgroup$ – Crowley Oct 19 '16 at 15:51
  • 16
    $\begingroup$ @Crowley yes, but it's even more obvious that the sum diverges, at the expense of the series being more complicated to describe $\endgroup$ – Richard Rast Oct 19 '16 at 17:41
  • 3
    $\begingroup$ I have to say this is the most beatiful divergent series I have ever seen. It's much easier to understand than the 1/2^n * 2^n one. $\endgroup$ – HopefullyHelpful Oct 20 '16 at 7:39
  • 4
    $\begingroup$ @HopefullyHelpful What about $1+2+3+4+5+\dots=-\frac1{12}$? $\trollface$ $\endgroup$ – PyRulez Oct 20 '16 at 12:41
  • 4
    $\begingroup$ @PyRulez It's very annoying that no-one ever seems to give, with that claim, the attendant context "The zeta-regularized sum $\sum\limits_{n=1}^{\infty} n^{-s}$ equals $\zeta(s)$, the Riemann zeta function. The Riemann zeta function's analytical continuation to $\zeta(-1)$ is $-\frac{1}{12}$". Moreover, people conveniently forget that these regularized divergent-series summations are not linear and stable; For instance, adding 0s can completely change the value of the series, as can reordering the terms of their summation. $\endgroup$ – Iwillnotexist Idonotexist Oct 22 '16 at 6:21
41
$\begingroup$

Do you think the series

$$1+\frac12+\frac12 + \frac14+\frac14+\frac14+\frac14 + \frac18 + \cdots$$

converges? Note, there are $2$ terms equal to $\frac12$, $4$ terms equal to $\frac14$, $8$ terms equal to $\frac18$ and so on, with $2^i$ terms equal to $\frac{1}{2^i}$ for each $i\in\mathbb N$.

You will probably agree that this series diverges. In fact, if you give me a number $m\in \mathbb N$, I can calculate exactly how many terms of the series you have to add together for the sum to reach $m$.

For example, it takes $1$ term to reach $1$, it takes $1+2=3$ terms to reach $2$, and then $1+2+4=7$ terms to reach $3$. You can show, with a simple inductive argument, that you will reach $m$ after

$$1+2+4+\dots + 2^{m-1}$$

terms, which is actually equal to $2^m-1$ and is certainly a finite number.


It's good to understand the concept why this series diverges. The thing is that yes, the terms go to $0$, but they don't do so "fast enough". The problem is than once the terms hit $\frac14$, they stick at that number for $4$ steps, long enough for the sum to increase by $1$.

And imagine what happens way way way down the line. The sum is equal to $\frac{1}{1024}$ for a whole $1024$ terms, for example. Sure, it will eventually fall to an even lower number, but it will stay on that number for even longer, again long enough for the whole sum to increase by $1$.


Side fact: the series I wrote down at the start has the bonus property that each term in the sequence is larger than the corresponding term of the sequence

$$1+\frac12+\frac13+\frac14+\frac15+\cdots$$

which is also known as the harmonic series and is the most famous divergent series. So, you now see that if you sum $2^m$ terms of the harmonic series, your sum will be equal to at least $m$ (and more, in fact).

$\endgroup$
  • 1
    $\begingroup$ "each term in the sequence is larger..." i think you mean each term of your series is less than or equal to each term in the harmonic series. $\endgroup$ – Steve Cox Oct 19 '16 at 15:18
  • 4
    $\begingroup$ @SteveCox - No. That part is correct. The error is in the statement about the sums of the harmonic series, which will be less than this series, not greater. $\endgroup$ – Paul Sinclair Oct 19 '16 at 15:29
  • 1
    $\begingroup$ @PaulSinclair No, actually Steve Cox was right. $\endgroup$ – 5xum Oct 20 '16 at 6:37
  • 4
    $\begingroup$ @5xum. You'd better check again. The 3rd term of your series is $1\over 2$, the third term of the harmonic series is $1\over 3$. ${1\over 2} > {1\over 3}$. For for the 5th, 6th, 7th terms, ${1\over 4} > {1\over 5}, {1\over 6}, {1\over 7}$, etc. Divide the terms in your series by $2$ and it works, but as stated, it does not. $\endgroup$ – Paul Sinclair Oct 20 '16 at 16:57
  • 1
    $\begingroup$ As an additional sanity check, the sum of the first $n$ terms of the Harmonic series is asymptotically $\ln n$, so the first $2^m$ terms sum to about $m \ln 2$, which is less than $m$ (but more than $m/2$). $\endgroup$ – Shagnik Oct 20 '16 at 19:19
14
$\begingroup$

However note that in the field of $p$-adic numbers $\mathbf Q_p$, a series $\sum_na_n$ converges if and only if the sequence $(a_n)$ tends to $0$.

$\endgroup$
12
$\begingroup$

You are echoing a 2-millenia-old quesite: will Achilles overtake the Tortoise ? , and
if Achilles decelerates,... Convergence/divergence of infinite sequences and series took long time to be mastered somehow, also because it was clear that the result of adding many small terms was depending on how many vs. how small.
Just consider Stereographic Projection and the philosophical debates it arouses.

$\endgroup$
  • $\begingroup$ in achilles and the tortoise, the problem is to demonstrate that the series converges $\endgroup$ – njzk2 Oct 19 '16 at 20:10
  • 2
    $\begingroup$ @njzk2 yes, but the question posed (and not well-answered by Ancient Greek mathematics) was "how do we correctly handle infinite series" $\endgroup$ – AakashM Oct 20 '16 at 9:28
  • 1
    $\begingroup$ Also, this illustrates why we care about rigorous definitions in math. Once we've defined what a sequence is, what convergence means, etc., this isn't a philosophical question; it has a definite answer. (Modulo mapping a question about a race in the real world to one about infinite series, but whatever.) $\endgroup$ – anomaly Oct 20 '16 at 16:36
  • $\begingroup$ @anomaly, yes actually I made a mixture, but I was mainly addressing the philosophical / psychological angst signalled by OP: how can "many" "vanishing" entities add up to infinite $\endgroup$ – G Cab Oct 20 '16 at 20:46
11
$\begingroup$

I will assume that Fundamental theorem of calculus is believable enough.

While I could relate to you that ratio or root tests are not that easy to understand intuitively, it doesn't get more visual than integral test for convergence. I mean, look at this. You can immediately see that $$\int_1^{n+1}\frac{dx}x\leq\sum_{k=1}^n\frac 1 k$$ Employ Fundamental theorem to get $$\ln(n+1)\leq\sum_{k=1}^n\frac 1 k$$ so, harmonic series diverges because logarithm is unbounded. And why is logarithm unbounded? Because it is the inverse of $\exp\colon \mathbb R\to \mathbb R_+$.

$\endgroup$
10
$\begingroup$

Just look at the problem from the other end.

Take a function like $f(x) = \sqrt{x}$ or $\log x$ - they grow infinitely with the $x$ growth, though their growth ratio decreases (their first derivatives tend to zero).

Consider the function increments for equal steps of argument, say, for arguments natural: $$\Delta f = f(n+1) - f(n)\quad \text{ for } n\in\mathbb N^+$$ As the $f$ function grows slower and slower, these deltas get smaller and smaller for growing $n$, and they actually tend to zero: for arbitrarily small $\varepsilon > 0$ there exists $n$ big enough, for which $\Delta f < \varepsilon$.

Now $\Delta f$ defines a sequence convergent to zero, whose partial sums are values of $\sqrt n$, that is a function growing with no bound, hence a divergent series.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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