Let's sum the reciprocals of squares.
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There's a few different approaches to trying to determine whether or not this
series converges or diverges. So here's the series I'm interested in.
The sum, n goes from 1 to infinity, of 1 over n squared.
And that series converges if and only if the series
n starts at 2 to infinity of 1over n squared converges.
All right, whether or not I include the end equals
1 term doesn't affect the convergence of the series at all.
So let's try to analyse this series now.
Well, note that 0 is less than or equal to 1 over n
squared is less than or equal to 1 over n squared minus n.
I mean, at least as long as n is 2 or more.
I don't want to plug in n equals 1 here, because then I'd be dividing
by 0. But this is true because this denominator
is smaller than this denominator so this fraction is bigger than this fraction.
Now what does that mean about the series. Well i can do a comparison test then.
So by comparison test.
The sum of 1 over n squared minus n, n goes from
2 to infinity converges. Implies that the smaller sum,
the sum n goes from 2 to infinity of 1 over n squared convergences.
So now I want to analyse the sum of 1 over n squared minus n.
Well, it's going to turn out that this series telescopes.
Let's see how that goes.
So, the series that I'm interested in analysing is, the sum
n goes from 2 to infinity of 1 over n squared minus n.
Al right, I'd like to evaluate that series.
Now, how could I do that?
The trick is to notice the following. What's 1 over n minus 1 minus 1 over n?
I can put that over a common denominator. So this is n over n times n minus
1 minus n minus 1 over n times n minus 1.
And now that I've got it over a common denominator, I can do the subtraction.
This is n minus n minus 1 over n times n minus 1.
Well, n minus n minus 1, that's just 1, and
the denominator is n-squared minus n, n squared minus n.
So, evaluating this series is really the same as evaluating
1 over n minus 1, minus 1 over n, the sum of these.
And before plunging into the infinity on top, let's
just do this from 2 to some value big n.
So what
happens when I plug in 2, I get 1 over 2 minus 1 is 1 minus 1 over 2.
And then I plug in n equals 3 I get 1 over 3 minus 1 which is 2 minus 1 over 3.
And then I plug in n equals 4 and then I get
1 over 4 and I get 1 over 3 minus 1 over 4.
And then I keep on going until I plug in n and I get 1 over
big N minus 1 minus 1 over n. But lots and lots of stuff cancels.
This is exactly what I mean by telescoping, right?
What cancels is minus and a 1 2 and this a 1 2, this minus a
1 3 and this a 1 3, this minus a 1 4 will cancel something in there.
Everything else in the middle will die.
There'll be a 1 over N minus 1 term, with
a negative sign in front of it, which'll cancel this.
The only thing that survives is this initial term,
1 over 1, and this last term, minus 1 over N.
So the sum of this, little n from 2 to big N, is this.
Now how do I evaluate the infinite series?
Well I just take a limit as big N goes to infinity.
So the limit as big N goes to infinity of the sum little n goes from 2 to big N
of 1 over little n minus 1 minus 1 over n. Is the limit as
big N goes to infinity of 1 over 1 minus 1 over big N.
Well, as big N goes to infinity 1 over N this term is going to 0.
So, it's 1 over 1 minus something very close to 0.
This limit is 1.
And that means, that this original series not only converges but I know its value.
Its value at 1.
Why is that significant? Well,
knowing that this series converges, then means that this series converges.
And if this series, little n from 2 to infinity of 1 over n squared converge.
That means that the original series that I'm interested in,
the sum of the reciprocals of the squares, that converges.
And that's great.
But that's not the only way to determine that this series converges.
So, yeah, I want to apply Cauchy Condensation, but to what series?
And I'm still working on this series, the sum of
1 over n squared, n goes from 1 to infinity.
But instead of looking at that series, I'm going to write it like this.
The sum of just the a sub n's where a sub n is 1 over n squared.
You know, I want to note, what, about this sequence a sub n?
It's a sequence who's terms are positive and it's
a decreasing sequence.
So it's this sort of thing that I'm allowed to apply Cauchy Condensation to.
And what does Cauchy Condensation say?
Well, it says that this series converges
if and only if the condensed series converges.
If and only if the sum of 2 to the n times
the 2 to the nth term, n goes from zero to infinity, converges.
So what's the condensed series?
In this case, just asking
whats this series in this case. Well that's the sum n goes from
0 to infinity of 2 to the n times what's the 2 to the n term of this sequence?
its 1 over instead of n, am going to write 2 to the n squared.
But I can simplify this, this is something times 1 over the same thing squared.
Well that's the sum n goes from 0 to infinity just
1 over 2 to the n. Does that series converge or diverge?
Well that series is just a geometric series, right?
So this series we've already seen, in this series converges.
And in fact we know its value, and it's value was 2.
And consequently, because this condensed series converges,
so too must the original series converge.
The sum of 1 over n squared as n goes from 1 to infinity converges.
So we've seen that this series converges by
comparing to a telescoping series that we know converges.
And by using Cauchy condensation, and there's other ways.
We could also have used the integral test.
So we're seeing lots of different methods to prove the same result.
The sum of one over n squared, n goes from 1 to infinity converges.
And usually were happy with that, usually were
happy just knowing that a series converges or diverges.
But in this case we can ask the more refined question.
To what does this convergent series converge?
So numerically, right the sum of the first two terms.
1 over 1 squared plus 1 over 2 squared, is 5 4.
And we can keep on going.
The sum of the first 3 terms, sum of the first
4 terms, 5 terms, 6 terms, 7 terms, 8 terms, 9 terms.
Little bit over 1.5. We could
try something a lot more terms, too, right?
Here's the sum of the first 100 terms, about
1.63, sum of the first 200 terms, first 300, 400.
And then if we add up the first 1000 terms, right?
So 1 over 1 plus 1 over 2 squared plus 1
over 3 squared da, da, da plus 1 over 1000 squared.
And we're getting just above 1.64.
The exact
answer is really pretty surprising.
Numerically right we're getting about 1.64 after adding about the first 1000 terms.
Pie squared over 6 is about 1.64 and it turns out that
the sum of the reciprocals of the squares is pie squared over 6.
This series evaluates to pie squared over 6 is a shocking result.
It should
really leave you wondering why, why is something like that true?
And unfortunately we have to wonder just a little bit longer.
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