General Question about Convergence

[nfcfox]

Here's what I'm confused about: The harmonic series 1/n is divergent (because you're just infinitely adding numbers) but the series (1/2)^(n+1) is convergent. This doesn't make sense to me because by plugging in larger and larger numbers, you are still adding small numbers infinitely which should make it divergent. I understand that for smaller values of n, (1/2)^(n+1) will have a smaller calculated number but you are still adding small terms up infinitely. Can someone explain the difference to me?

 

[.Scott]

The harmonic series 1/n is divergent (because you're just infinitely adding numbers)

That's not precisely correct. It's divergent because the series does not converge to a finite sum. This can be demonstrated by grouping the terms in increasingly longer groups. Those groups are [1/1], [1/2,1/3], [1/4,1/5,1/6,1/7], [1/8,...1/15], the next 16, th next 32, ... Each of these groups adds up to more than 1/2, there are an infinite number of these groups, so there is no upper bound to the total.

but the series (1/2)^(n+1) is convergent.

It is convergent because there is an upper bound to the sum. Each term brings it half way to the value of 1.

 

[nfcfox]

What do you mean by an upper bound? If I had an infinite number of groups like [1/2], [1/2,1/4], [1/2,1/4,1/8] and so on, it all adds up to above 1/2 too... I guess I'm confused on what convergence/divergence means with summations overall.

 

[Ray Vickson]

Here's what I'm confused about: The harmonic series 1/n is divergent (because you're just infinitely adding numbers) but the series (1/2)^(n+1) is convergent. This doesn't make sense to me because by plugging in larger and larger numbers, you are still adding small numbers infinitely which should make it divergent. I understand that for smaller values of n, (1/2)^(n+1) will have a smaller calculated number but you are still adding small terms up infinitely. Can someone explain the difference to me?

The DEFINITION of convergence of an infinite series $\sum_{n=1}^{\infty} x_n$ is that the finite sums $S_N = \sum_{n=1}^N x_n$ have a finite limit as $N \to \infty$. In the case of $\sum 1/n$ we can, for any finite large number $L>>0$ find an $N$ that gives $\sum_{n=1}^N 1/n = 1 + 1/2 + 1/3 + \cdots + 1/N > L$. So, we can have finite sums > 1 million, still larger finite finite sums > 100,000,000 trillion, etc. No matter how large a number $L$ you specify, you can make the finite sum be larger than that. The case of $\sum 1/2^{n+1} = (1/2) 1/2^n$ is different: you can actually compute the finite sums, using your high-school formulas for sums of geometric series: $\sum_{n=1}^N 1/2^n = (1/2 - 1/2^{N+1})/(1 - 1/2) = 1 - 1/2^N$. You can see right away that this has a finite limit (=1) as $N \to \infty$.

 

[PeroK]

What do you mean by an upper bound? If I had an infinite number of groups like [1/2], [1/2,1/4], [1/2,1/4,1/8] and so on, it all adds up to above 1/2 too... I guess I'm confused on what convergence/divergence means with summations overall.

It might help if you say what level your mathematics is at. Are you studying real analysis or have you just picked this up about convergence/divergence at random?

 

[nfcfox]

It might help if you say what level your mathematics is at. Are you studying real analysis or have you just picked this up about convergence/divergence at random?

I'm taking AP Calc BC as a senior in high school. I have a test tomorrow on sequences and series and I've been really just pulling along...

 

[PeroK]

I'm taking AP Calc BC as a senior in high school. I have a test tomorrow on sequences and series and I've been really just pulling along...

Try this in a calculator, spreadsheet or computer programme:

Take the series 1/2 + 1/3 + 1/4 + 1/5 + ...

and keep going until you get past 2. And, if you have enough patience, keep going unil you get past 3. And maybe past 4.

Now take the series 1/2 + 1/4 + 1/8 + 1/16 + ...

And keep going until you get past 2. Edit: or past 1, even!

 

[jbriggs444]

What do you mean by an upper bound? If I had an infinite number of groups like [1/2], [1/2,1/4], [1/2,1/4,1/8] and so on, it all adds up to above 1/2 too... I guess I'm confused on what convergence/divergence means with summations overall.

Note that your groups are not disjoint. You've counted the 1/2 term three times so far. If you add up the subtotals of each of your groups, that's not a fair way to try to sum up the total of the whole sequence.

By contrast, look at the groups proposed for 1/2 + 1/3 + 1/4 + ... Those are all disjoint. Every number gets counted in some group and no number gets counted in more than one group. That's a fair way to try to sum up the total of the whole sequence.

 

[SammyS]

You have been given some excellent replies here. I'll repeat the idea in post #2 in a little different way and in a little more detail for Σ(1/n) , then take a somewhat different look at Σ1/(2^(n+1)).

For Σ(1/n) :
$\displaystyle \sum_{n=1}^{\infty} =1+\frac12+\frac13+\frac14+\frac15+\frac16+\frac17+\frac18+\frac19+\frac1{10}+\frac1{11}+\frac1{12}+\frac1{13}+\frac1{14}+\frac1{15}+\frac1{16}+\frac1{17}+\frac1{18}+\frac1{19}+\dots \$

$\displaystyle =1+\frac12+\left(\frac13+\frac14\right)+\left(\frac15+\frac16+\frac17+\frac18\right)+\left(\frac19+\frac1{10}+\dots+\frac1{15}+\frac1{16}\right)+\left(\frac1{17}+\frac1{18}+\dots +\frac1{32}\right)+\dots \$​

.
The sum of the first two terms is $\ 1\frac12\ .$ Each of the next two terms is at least as great as 1/4, so their sum is a little greater than 1/2. Similarly the next 4 terms also sum to greater than 1/2, and so on.

No matter how large a number you might choose, it's possible to come up with a number of terms of this series which will have a sum larger than the chosen number.

For Σ1/(2^(n+1)) :
The first term is 1/2. The second is half of that and when added to 1/2 gives 3/4, which is 1/4 short of being 1 . The next term is half of 1/4, so it gets you half way to 1, still being short by 1/8. Adding the next term gets you half way to 1 again. This continues. Not only will the sum never get to 1, it will exceed any number less than 1, if enough terms are included.