Consider the series and its associated sequence of partial sums . Here we will assume that the numbers we are about to add are positive, that is, for any . It is clear that the process of generating the partial sums will lead to an increasing sequence, that is,

,

for any . Our previous knowledge about increasing sequences implies the following fundamental result:

**The positive series is convergent, if and only if, the sequence of partial sums is bounded; that is, there exists a number***M*> 0 such that,

for any .

**The Basic Comparison Test**):

- Consider the positive series .
**1.**- Assume there exists a convergent series such that,

then the series is convergent. **2.**- Assume there exists a divergent series such that,

then the series is divergent.

,

which means that the series is divergent.

**Example:**Show that the series

is divergent.

**Answer:**We have , for any ; hence

.

Since is divergent, we deduce from the Basic Comparison Test, that is divergent.

**Example:**Show that the series

is convergent.

**Answer:**Since

and the geometric series is convergent, then the series is convergent (using the Basic Comparison Test).

The next result (known as

**The p-Test**) is as fundamental as the previous ones. Usually we combine it with the previous ones or new ones to get the desired conclusion.

Consider the positive series (called the

**p-series**) . Since the limit of the numbers must add to 0, in order to expect convergence, we assume that

*p*> 0. The next result deals with convergence or divergence of the series

when

*p*>0.

**The p-series converges, if and only if, .**

- Consider the function defined by.

It is easy to check that*f*(*x*) is decreasing on . Hence, for any , we have for any ,

which implies

,

that is,

.

Using this inequality, we get

and

.

If , then we have

and if*p*=1, then we have

.

**Case 1:***p*< 1, then we have.

Since , then the series is not bounded,and therefore it is divergent.**Case 2:***p*> 1, then we have

but, since

,

we get

,

which means that the sequence of partial sums associated to the series is bounded. Therefore, the series is convergent.

**Case 3:***p*= 1, we have already shown that the series is divergent. Though one may want to easily check that we have,

which shows that the sequence of partial sums is not bounded.

**Example:**Discuss the convergence or divergence of

.

**Answer:**It is not hard to show that for any , we have . Then, we have

.

Since, by the p-Test, the series is convergent, the Basic comparison Test implies that is convergent.

The last result on positive series may be the most useful of all. Indeed, the

**Limit Test**should be always in mind when it comes to cleaning up some undesirable terms.

Before we state this test, we need a new notation. Indeed, we will say that the two sequences and are equivalent, or , if and only if,

.

**Let and be two positive series such that.**

Then converges, if and only if, converges.

**Example:**Determine whether the series

is convergent or not.

**Answer:**Note that when

*n*is large we have and . Then it is easy to check that

.

Using the p-test we get that the series is convergent. Hence, by the Limit-test, we deduce the convergence of the series .

**Example:**Determine whether the series

is convergent or not.

**Answer:**We know that

.

This limit means that when

*x*is very small , then (a very useful conclusion in physics, for example, when dealing the motion of the pendulum). So when

*n*is large, 1/

*n*will be small and therefore . This clearly implies that

Since the series is divergent, the limit-test implies that the series is divergent.

**Remark:**It should be appreciated that, without the Limit-test, it would be very hard to check the convergence.