∞

1
The harmonic series ∑ diverges.
n =1 n
1 1 1 1 1 1 1 1
Let sn be the nth partial sum of the series, i.e., sn=1+ + + + + + + + … .
2 3 4 5 6 7 8 n
1
Here, s2=1+ .
2

( )
1 1 1
2 3 4
1 1 1
2 4 4
2
s4 =1+ + + >1+ + + =1+ .
2 ( )
2
Therefore, s2 >1+ .
2
2
1 1 1 1 1 1 1
Again, s8 =1+ + + + + + + =1+ +
2 3 4 5 6 7 8
1
2 ( 13 + 14 )+( 15 + 61 + 17 + 18 )
1
or, s2 >1+ +
3
2 ( 14 + 14 )+( 18 + 18 + 18 + 18 )
1 1 1
or, s2 >1+ + +
3
2 2 2
3
or, s2 >1+ .
3
2
 n
In general, s2 >1+ .
n
2

Now, lim s 2 > lim 1+

Definition:

The series ∑ an is said to converge absolutely if the series ∑ |an| converges.

Theorem: If ∑ an converges absolutely, then ∑ an converges.

Proof:

Let ∑ an be a series, and suppose it converges absolutely. This means that the series ∑ |an|
converges. We need to show that the series ∑ an converges.

Since ∑ |an| converges, by definition of convergence, for every ϵ >0 , there exists an integer

| ∑ | ||
m
N such that for all m>n ≥ N , we have a k < ε.
k=n +1

m
Now, consider the series ∑ an . We need to show that the partial sums Sm =∑ a k form a
k=1

Cauchy sequence. That is, for every ϵ >0 , there exists an integer N such that for all m>n ≥ N ,
we have |S m−S n|< ε .
m
Note that Sm −S n= ∑ ak .
k=n +1

|∑ |
m m
Clearly, ak ≤ ∑ |ak|
k=n +1 k=n+ 1
 m
Since ∑ |an| converges, we can make ∑ |a k| arbitrarily small by choosing nand m large
k=n +1

enough. Specifically, for the given ϵ >0 , there exists N such that for all m>n ≥ N ,
m

∑ |a k|< ε .
k=n +1

|∑ |
m m
Therefore, ak ≤ ∑ |ak|< ε .
k=n +1 k=n+ 1

This shows that the partial sums Sm form a Cauchy sequence. Since R (the set of real
numbers) is complete, every Cauchy sequence converges to a limit. Hence, the series ∑ an .
converges.
This completes the proof.

Example 1:
1
Test the convergence of the alternating harmonic series: ∑ an=∑ (−1)n n .
Absolute Convergence
First, let us test for absolute convergence:

Here, |
∑|an|=∑ (−1)n 1n =∑ 1n | , which is a harmonic series. We know that the harmonic

1 1
series diverges. Therefore, ∑|an|=∑ n diverges, meaning ∑ (−1)n n does not converge

absolutely.
Conditional Convergence
Next, we test for conditional convergence using the Alternating Series Test. The test states
that an alternating series ∑ (−1)n b n converges if:

1. b n is monotonically decreasing and

2. nlim bn =0.
→∞

1
For our series ∑ (−1)
n
:
n

1 1 1
1. b n= is monotonically decreasing because < for n ≥ 1.
n n+1 n
1
2. lim =0.
n→∞ n
 1
Since both conditions are satisfied, the alternating series ∑ (−1)
n
converges.
n
Conclusion:
1
The series ∑ (−1)n n does not converge absolutely, but it does converge conditionally.

1
Therefore, the alternating harmonic series ∑ (−1)
n
converges.
n

Example 2:
1
The series ∑ an=∑ (−1)n n2 converges absolutely.
1
To show that the series ∑ (−1)
n
2 converges absolutely, we need to check the convergence
n
of the series of its absolute values:

| 1
∑|an|=∑ (−1 )n n2 =∑ n2 . | 1

1 1
The series ∑ n2 is a p-series with p=2. A p-series ∑ np converges if and only if p>1 .

1
Since p=2> 1, the series ∑ 2 converges.
n

1
Since the series of absolute values converges, the original series ∑ (−1)n n2 converges

absolutely.

Alternating series
Theorem
Suppose
(a) |c1|≥|c 2| ≥|c3|≥|c 4| … ,
(b) c 2 m−1 ≥ 0 , and c 2 m ≤0 , m=1 , 2 ,3 , … ;

(c) nlim c n=0 .

Then ∑ c n converges.

Proof: |c1|≥|c 2| ≥|c3|≥|c 4| … indicates that the absolute values of the terms are non-increasing,
whereas c 2 m−1 ≥ 0 , and c 2 m ≤0 , m=1 , 2, 3, …implies that the series alternates in sign.
Define b n=|c n|. Since the absolute values |cn| are non-increasing, we have b n ≥ bn +1. Given

that nlim c n=0 , it follows that lim bn =0.

Now we consider the series ∑ (−1)n−1 b n. By construction, we have b n ≥ bn +1 and nlim bn =0.
→∞

Therefore, by the Alternating Series Test, the series ∑ (−1)n−1 b n converges.

Epsilon-delta definition of a limit

Example: lim (3 x +1)=7.

Solution

To prove this using the epsilon-delta definition, we need to show that for every ε > 0, there

exists a δ >0 such that whenever 0<∣ x−2∣< δ , it follows that ∣(3 x+1)−7 ∣<ε .
∣(3 x+1)−7 ∣<ε , which on simplification gives ∣3 ( x−2)∣< ε , i.e., ∣ x−2 ∣<ε /3.
Given ε¿ 0 , let δ=ε /3.
If 0<∣ x−2∣< δ , then 0<∣ x−2∣< ε /3, i.e., ∣3 (x−2)∣< ε , implying that ∣(3 x+1)−7 ∣<ε .

Thus, lim (3 x +1)=7.