Series CheatSheet

1 Definitions and properties
Series CheatSheet
Definition 1.1. Let an be a sequence of real numbers. Denote the partial sum of the first n
elements in the sequence with Sn:
n
Sn =
The function that maps n to Sn is a real sequence and we call it a series. If the sequence is
convergent we denote the limit with:
+∞
i=1
Proposition 1.2. Let an be a sequence of non negative numbers. If the series Sn = n
i=1 ai is
bounded from above it’s convergent.
Proof. If all the summands an are positive, the sequence Sn is increasing. An increasing sequence
which is bounded from above is convergent.
Proposition 1.3. Let an be a sequence and Sn be the associated series. If Sn is convergent, the
sequence an converges to zero.
Remark 1.4. The implication works in one direction only!!!! There are plenty of sequences
an that converge to zero, but their associated series are divergent. For example an = 1
n .
2 Techniques to study the convergence
1st method: The following technique is useful to determine the convergence of some very basic
series. We will call it the 2 k -method.
i=1
ai
Proposition 2.1. Let an be a positive and decreasing sequence. The series +∞
n=1 an is
convergent if and only if the series:
+∞
2 k a2k is convergent.
k=0
2nd method: The following technique lets us compare a series with an improper integral. Sometimes
it might be easier to evaluate an improper integral than a series (most of the time this
is not the case):
1
ai

Proposition 2.2. Let f(x) be a real function on the interval [1, +∞) which is positive,
continuous and decreasing. Let n be a natural number, the following inequality holds:
n+1
1
f(x) dx ≤
n
f(i) ≤
i=1
n+1
1
f(x) dx + f(1) − f(n)
The series +∞
i=1 f(i) is convergent if and only if the improper integral +∞
1 f(x) dx is convergent.
3rd method: With the following proposition we can compare a given series with some other series
that we have already studied:
Proposition 2.3. Let an, bn be two positive sequences. Suppose that there is a natural number
n such that for all n > n the following inequality holds:
an ≤ bn
Under these assumptions if the series +∞
n=1 bn is convergent, the series +∞
n=1 an is also
convergent. If the series +∞
n=1 an is divergent, the series +∞
n=1 bn is also divergent.
4th method: The following technique is the basic tool to turn a complicated series into something
simpler. It’s not the ultimate technique (there are cases where it’s inconclusive) but it’s the
sharpest tool that we have (for the moment).
Proposition 2.4. Let an, bn be positive sequences. Suppose that the sequence bn converges to
a real number C > 0. The series +∞
n=1 anbn is convergent if and only if the series +∞
n=1 an
is convergent.
If the limit of bn is zero we have a weaker statement:
Proposition 2.5. Let an, bn be positive sequences. Suppose that the sequence bn converges
to zero. If the series +∞
n=1 an is convergent, the series +∞
n=1 anbn is also convergent.
3 Fundamental limits
The following limits of sequences are fundamental. These should be used together with the limit
comparison technique.
•
•
•
•
lim
n→+∞
lim
n→+∞
lim
n→+∞
sin 1
n
1
n
1 − cos 1
n
1
n 2
= 1
= 1
2
ln(1 + 1
n )
1 = 1
n
e
lim
n→+∞
1
n − 1
1 = 1
n
2

4 Strategy
There is no algorithm to determine the convergence of a series, nonetheless we can try to outline a
strategy that would work in many circumstances.
Step 1: If the sequence an doesn’t converge to zero the series is divergent and we are done.
Step 2: If the limit is zero the series can be convergent divergent or might not have a limit.
If possible, we apply limit comparison repeatedly to replace transcendental functions
with rational functions. We replace in the following way:
•
•
•
•
sin 1 1
≈
n n
cos 1 1
≈ 1 −
n 2n2 ln(1 + 1 1
) ≈
n n
e 1
n ≈ 1 + 1
n
If after replacing the transcendental pieces the sequence is equal to zero, limit comparison
is inconclusive!!!
Step 3: If the sequence is an algebraic function (polynomials, divisions and radicals only) we
collect the highest power of n and we apply limit comparison again.
Step 4: If everything worked fine the series should be a fundamental series (one in the list
below) or more in general a product/quotient of the following functions: n p , a n , ln q (n), n!, n n . At
this point we can try to conclude the calculation using the integral test, the 2 k -test or comparison.
Example 4.1. Determine for which positive real numbers α, β the following series is convergent:
+∞
n=1
1
sin
nβ
ln α (n 2 + 1)
We apply limit comparison and replace sin and we obtain the following equivalent series:
+∞
n=1
We extract the leading term from the logarithm:
+∞
n=1
1
n β lnα (n 2 + 1)
1
nβ
2 ln(n) + ln 1 + 1
n2 α We don’t need to approximate ln 1 + 1
n2
since we are summing it to ln(n) which is divergent. We
apply limit comparison again:
+∞ (ln(n)) α
n=2
3
n β

This series cannot be made any simpler. We can try to apply the 2 k -test and we obtain:
+∞
n=1
n α
2 n(1−β)
We conclude by comparing with the improper integral:
+∞
It becomes more familiar if we replace 2 with e:
+∞
1
1
x α 2 x(1−β) dx
x α e x(1−β) ln(2) dx
If β − 1 > 0 we can make the substitution t = x(β − 1) ln(2) and we obtain:
+∞
x
(β−1) ln(2)
α e −x dx
which is convergent for every value of α. The series is convergent for β > 1 and α > 0 and not
convergent for 0 < β ≤ 1 and any value of α. We could have skipped the 2 k -test and apply the
integral test directly (this second approach is not any simpler).
5 Fundamental series
The following series should be considered fundamental, and you don’t have to retrieve them each
time you do a calculation:
•
•
•
•
+∞
n=1
It is convergent if and only if p > 1. It can be retrieved with the integral test or the 2 k -test.
It is convergent if and only if |p| < 1. It can be calculated explicitly.
+∞
n=1
+∞
n=1
1
n p
p n
1
n(ln(n)) p
It is convergent if and only if p > 1. It can be retrieved with the integral test or the 2 k -test.
+∞
n=1
It is convergent. It can be done by comparison.
4
n!
n n

•
•
•
It is convergent. Done by comparison.
It is convergent for every p. It can be retrieved with the integral test.
+∞
n=1
+∞
n=1
+∞
n=1
e n
n!
n p
e n
ln(n)
n p
It is convergent if and only if p > 1. It can be retrieved with the integral test or the 2 k -test.
The number p in all the previous formulas is a real number.
6 Good but not optimal estimate of the factorial
Sometimes it’s useful to compare the factorial with some better looking function. We start from
the identity:
n
ln(n!) = ln(i)
Since ln is monotonic (increasing) we can compare with the integral:
n
1
ln(x) dx ≤
i=1
n
ln(i) ≤
i=1
n+1
1
ln(x) dx
We can integrate the logarithm and we have:
n
n ln(n) − n + 1 ≤ ln(i) ≤ (n + 1) ln(n + 1) − n
We can now remove the logarithm by exponentiating the inequality:
i=1
nn (n + 1)n+1
≤ n! ≤
en−1 en Example 6.1. Determine if the series is convergent or divergent:
Applying the estimate we have:
Since the series +∞
n=1
well.
+∞
n=1
3 n n!
n n
3n en−1 ≤ 3nn! nn ≤ 3n (n + 1) n+1
nnen n 3
e is divergent, by comparison the series we are studying is divergent as
5

This estimate is good enough for many applications but it’s not optimal since:
lim
n→+∞
This means that n! is not approximated by
can be proven that:
lim
n→+∞
n!
n = +∞
n
e
n n
e . However, by some more advanced technique it
n!
√ 2πn
n
e
n = 1
This means that for large values of n the factorial is approximated by √ 2πn
as Stirling’s formula and it goes beyond the purpose of these notes.
6
n
e
n
. This is known