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In trying to understand why $\sum\limits_{k=1}^{\infty} \frac{1}{2^k}$ converges but $\sum\limits_{k=1}^{\infty} \frac{1}{2k}$ doesn't, I noticed that in infinite series of the type $\sum\limits_{n=1}^{\infty} \frac{1}{k^n}$ where $k > 1$, any term is greater than the sum of any number of subsequent terms.

Whereas for example in the series $\sum\limits_{k=1}^{\infty} \frac{1}{2k}$, it is always possible to find for any term a certain number of subsequent terms whose sum is greater than that term.

So I'm wondering, is there anything to this? Is this principle the difference between a series converging or diverging?

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    $\begingroup$ You'll find series with growth "between" those you list - like summing over $\frac{1}{n\log(n)}$ diverges, but not as quickly as $\frac{1}n$ - but summing over $\frac{1}{n\log(n)^2}$ converges, though not as quickly as $\frac{1}{k^n}$! Really, the line between convergent and divergent series is not as obvious as one might hope. $\endgroup$ – Milo Brandt Jun 10 '15 at 0:50
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Convergence of series always goes back to the definition with the limit of the partial sums. You have touched on some parts that are true. For example, you say (paraphrasing) that if one term, say $a_{1}$ is bigger than the sum of all the rest (for example when $a_{n}=\frac{1}{2^{n}}$) then the series converges. (Assuming all terms are positive) When this is true, you get convergence because the sum is bounded above:

$$\sum_{n=1}^{\infty} a_{n} = a_{1}+\sum_{n=2}^{\infty}a_{n}\leq 2a_{n}<\infty.$$

The reason this implies convergence is that the sequence of partial sums $\sum_{n=1}^{N}a_{n}$ is an increasing sequence (since all the terms are non-negative) and is bounded above (by $2a_{1}<\infty$), so it converges.

One thing you should be careful of though is that this isn't true whenever $a_{n}=\frac{1}{k^{n}}$ for some $k>1$. The series will converge, but $a_{1}\leq \sum_{n=2}^{\infty}a_{n}$ if $1<k<2$. What you are catching here is essentially the ratio test (when subsequent terms drop in size quickly the series converges).

On the other hand, series like $\sum_{n=1}^{\infty}\frac{1}{2n}$ do not converge because the terms that are being added don't go to zero fast enough, i.e. the change from $a_{n}$ to $a_{n+1}$ is too small.

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  • $\begingroup$ Thank you, this is helpful. Is there a formal definition or a way to understand "fast enough" in the context of "going to zero"? $\endgroup$ – jeremy radcliff Jun 10 '15 at 0:13
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    $\begingroup$ @jeremyradcliff, there's no "good" formal definition that I'm aware of. You just kind of get a feeling for it. For example $\frac{1}{n^{p}}$ goes to zero fast enough provided that $p>1$ and it isn't fast enough if $p\leq 1$. $\frac{1}{p^{n}}$ goes to zero fast enough provided that $p>1$ (exponential decay). $\frac{1}{n!}$ and $\frac{1}{n^{n}}$ both go to zero much faster. The more of these you look at, the more experience you get, and the better "intuition" you develop. $\endgroup$ – TravisJ Jun 10 '15 at 0:24
  • $\begingroup$ Thanks for the details. I just started learning about series and this stuff is fascinating to me. I get stuck very quickly though, trying to immediately "understand" why something would or wouldn't converge, so I'll just keep playing with different kinds of series and see how my Convergence radar evolves. $\endgroup$ – jeremy radcliff Jun 10 '15 at 0:28
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    $\begingroup$ @jeremyradcliff, I always really liked determining whether series converged or not because it seemed like there was no "one way to do it" and no "formula" that just always worked. I viewed them as little puzzles waiting to be solved. I still enjoy them... but now, most are either easy, or impossibly hard. $\endgroup$ – TravisJ Jun 10 '15 at 0:33
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"Is this principle the difference between a series converging or diverging?"

I don't think so. For example, just consider a geometric series $$\sum_{k=0}^\infty r^k$$ with $0<r<1$. This will always converge. However, $$\sum_{k=n+1}^\infty r^k=\frac{r^{n+1}}{1-r}\ ,$$ and now

  • if $r<\frac12$ then this is less than $r^k$: that is, any sum of subsequent terms is less than the $k$th term; whereas on the other hand
  • if $r>\frac12$ then this is greater than $r^k$: that is, there exists a sum of (finitely many) subsequent terms which is greater than the $k$th term.

So as far as I can see, the observation has, in general, nothing to do with convergence.

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  • $\begingroup$ Thank you, that's very clear. It's too bad, this principle somehow kind of made sense to me intuitively. Come to think of it, it did seem a little simple... $\endgroup$ – jeremy radcliff Jun 10 '15 at 0:21
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    $\begingroup$ If you haven't seen it already, you might be interested in looking up $p$-adic analysis. In this area, a series converges if and only if its $k$th term tends to zero, which makes it (in some respects) much easier than real analysis. (However it is, in effect, dealing with different numbers, so it isn't a replacement for real analysis.) $\endgroup$ – David Jun 10 '15 at 0:27
  • $\begingroup$ Thank you, I'll check it out. $\endgroup$ – jeremy radcliff Jun 10 '15 at 1:54
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Quick Reason: One is geometric and the other is a multiple of the harmonic series.

Using the Integral Test,

$$\sum_{k=1}^{\infty} \frac{1}{2k} \ \ \textrm{divg/convg} \iff \lim_{k \to \infty} \int_{1}^{k} \frac{1}{2x} \ \textrm{dx}\ \ \textrm{divg/convg}$$

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    $\begingroup$ Thank you, but I wasn't really asking about those two in particular; my question was (perhaps poorly phrased on my part) whether the principle I describe, whereby one can always or never find for any term in a series a subsequent number of terms whose sum is greater than that term, is generalizable as a condition for Convergence/Divergence. $\endgroup$ – jeremy radcliff Jun 10 '15 at 0:03
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    $\begingroup$ That is the general idea. What you are stating is the fact that a series will converge $\iff$ it's sequence of partial sums are bounded. The fact that you can always get a partial sum greater that a number is this unboundedness that makes the given series diverge. $\endgroup$ – Mr.Fry Jun 10 '15 at 0:05

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