Converge or Diverge?
Learning goal: Students start to use convergence tests—nth term and integral test are introduced.
Since series are really limits of partial sums, anything we know about limits we know about
∞
series. For instance, if
∑a
= A and
n
n=1
converge), then
∞
∑ (a
n=1
∞
∑b
n
= B (in other words, each series’ partial sums
b=1
c
c
⎛ c
⎞
+
b
)
=
lim
(a
+
b
)
=
lim
a
+
bn ⎟ . Now the limit of a sum is
∑
∑
∑
n
n
n
n
n
⎜
c→∞
c→∞ ⎝
⎠
n=1
n=1
n=1
c
c
the sum of the limits, so this breaks into lim ∑ an + lim ∑ bn which is, of course A + B. Smilarly
c→∞
n=1
c→∞
n=1
we could multiply sums by a constant.
It’s often important to know if a series converges or diverges before looking at it more carefully.
If it diverges, it’s not really worth analyzing further, while even if the partial sums aren’t
immediately clear, it might be worth looking further if we know for sure that the series
converged. Hence, mathematicians have built up a toolbox of tests to determine if series
converge or not, without having to know exact formulas for the partial sums.
One very simple test is this. For the partial sums to converge, they have to get and stay close
together. That means the number you add at each stage must get and stay close to zero.
The nth-term test for divergence: if the nth term does not go to zero, the series diverges. That
is, unless lim an = 0 the series diverges. This includes series where the terms don’t even have a
n→∞
∞
limit, e.g.
∑ sin(n) .
n=1
WARNING!!!! There is no such thing as an nth term test for convergence. The terms might go
to zero and the sum will diverge anyway. ANYONE who EVER writes “the terms go to zero, so
the series converges” will automatically get a zero on that problem. Period.
Example: 1/1 + 1/2 +1/3 + 1/4 + ! diverges. This is because 1/1 > 1/2, and 1/2 + 1/3 > 1/4 +
1/4 = 2/4 = 1/2, 1/4 + 1/5 + 1/6 + 1/7 > 1/8 + 1/8 + 1/8 + 1/8 = 4/8 = 1/2. And so on. Each
chunk of terms is > 1/2, so n chunks of terms adds up to greater than n/2. So the series cannot
converge because the partial sums blow up. Even though the terms go to zero, the series
diverges.
Now, we’ve seen how, structurally, infinite series look like improper integrals. Maybe each can
inform the other.
<<Integral Test Worksheet>>
So to recap the worksheet, if our series an can be though of as the values of a function a(n) which
is positive and decreasing, then for any fixed n, we have
∫
n
1
n
a(x) dx < ∑ ak (in fact the sum only
k=1
has to go through n – 1, not even n). In other words, the integral up to n is less than the nth
partial sum. But if we add the first term to the integral, it is now bigger than the sum. This holds
for any n, and therefore it holds in the limit (if the limits exist).
If the limits do not exist, then we know (because the function is positive) that the only way the
limit could not exist is if it blows up. But then the sum, being larger, also blows up. So if the
integral diverges, so does the sum.
On the other hand, if the integral converges, the sum is always increasing, but bounded above by
the integral-plus-first-term. So it, too, must converge. In other words, if the integral converges,
so goes the sum. Not only that, the sum is more than the integral, but less than the integral plus
the first term of the sum.
∞
1
dx
diverges
(p-type),
so
also diverges (as we have already seen). On the
∑
∫1 x
n=1 n
∞
∞ dx
1
other hand, by p-type, ∫ 2 converges to 1, so ∑ 2 also converges and is between 1 and 2.
1 x
n=1 n
It’s important to know this, because it’s impossible to find a nice formula for the partial sums of
this last sum. For a very long time, mathematician knew it converged, but not what it converged
to. Euler was the first person to figure out that it converges to π2/6. (Why? You’ll have to
wait!)
Example:
∞
More practice: (HH section 9.3, p.480) 1, 2, 5, 7, 9, 12, 13 – 33 (some), 35