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I'm trying to get the basic intuition on this without the use of any comparison or integral tests.

From what I understand, you have a series 1/n^p. If n <= 1 then as n tends the infinity, the partial sums of the terms will exceed the number of terms. So for very large n

Sn > n

But at the same time you'll keep adding terms that are approaching zero. (I'm assuming now 0 < p < 1 so 1/n^p decreases) and we know from the divergence test that when a series converges then a(n) tends to zero, and the series converge because you add very small values so at infinity the sum stops growing.

I fail to see the connection here. From a(n) -> 0 as n -> infinity which tells me the series should be converging. And from S(n) > n, the n'th partial sum is greater than n?

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    $\begingroup$ It may be easier to consider $1+{1\over2}+{1\over2}+{1\over3}+{1\over3}+{1\over3}+\cdots$. The terms converge to $0$, but the sum diverges, since "chunks" of it sum to $1$. $\endgroup$ – David Mitra Aug 15 '15 at 8:48
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    $\begingroup$ To compute a sum $\sum_{k=1}^\infty a_k$ with $a_k > 0$, there are two factors in competition. The terms $a_k$ getting smaller and you are adding more terms. Ultimately, what happens depend on which factor is more aggressive. If $a_k$ decreases fast enough, then the series converges. If not, the series diverges. The harmonic series $\sum_{n=1}^\infty\frac{1}{n}$ simply doesn't decrease fast enough. $\endgroup$ – achille hui Aug 15 '15 at 8:52
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    $\begingroup$ You certainly do not have $S_n > n$, for any $n$. Observe: $$S_n = 1 + \frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n} < \underbrace{1+ \cdots + 1}_{n \text{ summands}} = n$$ $\endgroup$ – 727 Aug 15 '15 at 8:55
  • $\begingroup$ "and the series converge because you add very small values..." . NO.... If the series converges the terms must tend to 0, but not the other way around. $\endgroup$ – DanielWainfleet Aug 15 '15 at 20:00
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Generally speaking $\sum_{k=1}^\infty \frac{1}{k^p}$ diverges for all $0< p \leq 1$, and converges absolutely for all $p > 1$.

It is clear that if mentioned series diverges for $p=1$, then it should diverge for $p \in (0;1)$ also.

Notice: \begin{align} \sum_{k=1}^\infty \frac{1}{k} = & 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} \dots \geq \\ & 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{16} + \dots = \\ & 1 + \sum_{k=1}^\infty \frac{1}{2} = \infty \text{.} \end{align}

Also if general term of a series tends to $0$ it does not necessarily mean, that the series converges, as you can see in the above example. But there is an implication: $$ \text{a series converges} \implies \text{general term tends to $0$} \text{.} $$

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  • $\begingroup$ But why does the implication not go the other way around, how is it possible that at very large n, you add terms that are approaching zero and yet the sum of the terms still grows? $\endgroup$ – Arnold Doveman Aug 15 '15 at 9:11
  • $\begingroup$ @ArnoldDoveman "how is it possible" - well, what's stopping it? $\endgroup$ – whacka Aug 15 '15 at 9:16
  • $\begingroup$ It is possible, that even though general term ($a_k$) tends to $0$, the series diverges, as you can see in the example I have just posted. Term $\frac{1}{k}$ approaches $0$ not fast enough. $\endgroup$ – Wojciech Karwacki Aug 15 '15 at 9:17
  • $\begingroup$ I understand now that the series diverges because the terms grow slowly, I just dont get why that is. Why does the rate at which the terms tend to zero have anything to do with it diverging, the bottom line is that the terms are decreasing, so at large n the terms are very small. And when you keep adding small values to the sum shouldnt the sum stop growing ? My apoligies, this is just very confusing to me $\endgroup$ – Arnold Doveman Aug 15 '15 at 9:26
  • $\begingroup$ Well this is just how the things work. Yes, you keep adding smaller and smaller values. But you end up adding infinite number of them and the fact that they approach $0$ slowly makes each consecutive summation bigger, but not bounded by some constant value. I'm pretty sure that if you understand how example in my answer works you will get pretty good feeling of what's going on with partial sum of this series. $\endgroup$ – Wojciech Karwacki Aug 15 '15 at 9:35
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To understand this in a very intuitive manner, you might want to see this.

This proof is what made me understand it both intuitively and rigorously.

$H_{2n}-H_{n}=\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2n}>\frac{1}{2n}+\frac{1}{2n}+\ldots+\frac{1}{2n}=n\cdot\frac{1}{2n}=\frac{1}{2}$

Thus, when you take sums from $n$ to $2n$, you're adding a term to $H_n$ that is bigger than $\frac{1}{2}$. However, you know you can continue this way forever, as in, from $n$ to $2n$ and from $2n$ to $4n$, etc.

Thus, you're always able to add a $\frac{1}{2}$ more, and thus, the series diverges.

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$a_n\to 0$ as $n\to\infty$ only tells you that the series might be converging because it is not already for trivial reasons divergent.

Also, we do not have that the $n$th partial sum $S_n$ is $>n$. The partial sums for the harmonic series grow much slower (but they still grow beyond any bound). For example, it takes more than $10^{400}$ summands until $S_n$ grows bigger than $1000$.

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  • $\begingroup$ Wouldnt that still mean, correct me if I'm wrong, that for largne n Sn > n. In your case when n = 10^400. Which is still not intuitive for me as how can the total sum of terms be greater than the number of terms when the terms you are adding are very small, as n tends to infinity. $\endgroup$ – Arnold Doveman Aug 15 '15 at 9:07
  • $\begingroup$ @ArnoldDoveman $S_n\to\infty$ does not mean $S_n>n$ for large $n$. For instance $\log(n)\to\infty$ but $\log(n)$ does not become bigger than $n$ for large $n$. $\endgroup$ – whacka Aug 15 '15 at 9:15
  • $\begingroup$ @ArnoldDoveman just think of $a_n=n-1$. That certainly tends to $\infty$ yet $a_n<n\forall n$ $\endgroup$ – Hasan Saad Aug 15 '15 at 10:48
  • $\begingroup$ @HasanSaad, when you say an are you refering to the sequence? Or Sn for the sum of the first n terms, the nth partial sum. $\endgroup$ – Arnold Doveman Aug 15 '15 at 10:59
  • $\begingroup$ The sequence. Obviously, however, we can write this as a series if you want. $s_n=a_{n}-a_{n-1}$. Again, the $n$th partial sum is $n-1<n$ yet tends to $\infty$. $\endgroup$ – Hasan Saad Aug 15 '15 at 11:26
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Perhaps recast the problem in a different light: instead of thinking about the sums, think about the sequence of partial sums. Say you are tasked with picking point after point on the number line, and each point you pick must be to the right of your previous choice. But you must make sure that the gaps between consecutive points are shrinking to nothing over time. So what happens?

One scenario: say you (mentally) pick an upper bound $U$, and then in your task you simply pick numbers that keep getting closer and closer to $U$ from the left. This describes convergence.

Different scenario: say you partition each interval $[n,n+1]$ (with natural numbers $n$) into $n$ pieces, so for instance $[2,3]$ is split into $[2,2.5]$ and $[2.5,3]$, and then $[3,4]$ is split up into the three intervals $[3,3\frac{1}{3}]$, $[3\frac{1}{3},3\frac{2}{3}]$ and $[3\frac{2}{3},4]$. Now take all of the endpoints of these intervals as your chosen points for the sequence of partial sums. Sketch this yourself on the number line to make sure you understand what's going on. Even though the gaps between consecutive points is shrinking to nil over time, the points themselves continue on to the right without bound! The fact that the gaps are getting smaller means our trek rightwards on the number line is going slower and slower with each point, but not slow enough to entail any upper bound in this case.

This is exactly the example David Mitra gave as the very first comment to this question:

$$1+\frac{1}{2}+\frac{1}{2}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\cdots $$

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