A test to decide whether the given series is divergent or convergent. Here, we consider two series and we will check how fast this series grow. We consider the limit here. If the limit is zero, then the term below (denominator) grows faster than the other. So, the series below converges. And so, the slow growing series in the top also converges.

When the limit is greater than zero, then both the series diverges or converges together. When the limit tends to infinity, the series below grow slowly. So, it is divergent. Hence, the series at the top is also divergent.

Two types of comparison tests shall now be discussed. In the first type, the general term of one series will be compared with the general term of the second series. In the second type, the ratio of two consecutive terms of one series will be compared to the ratio of the corresponding consecutive terms of the second series.
If $\sum u_{n}$ and $\sum v_{n}$ are two positive term series, and k$\neq$ 0, a fixed positive real number (independent of n) and there exists a positive integer m such that u$_{n}$ $\leq kv_{n}$, $\forall$ n $\geq$ m, then

1) $\sum u_{n}$ is convergent, if $\sum v_{n}$ is convergent, and

2) $\sum v_{n}$ is divergent, if $\sum u_{n}$ is divergent.

Proof: Let n $\geq$ m,

S$_{n}$ = u$_{1}$ + u$_{2}$ + ......... + u$_{n}$, and t$_{n}$ = v$_{1}$ + v$_{2}$ + ....... + v$_{n}$

Now $\forall$ n $\geq$ m, we have

S$_{n}$ - S$_{m}$ = u$_{m+1}$ + u$_{m+2}$ + .... + u$_{n}$

$\leq$ k(v$_{m+1}$ + v$_{m+2}$ + ........ + v$_{n}$) = k(t$_{n}$ - t$_{m}$)

S$_{n}$ $\leq$ kt$_{n}$ + (S$_{m}$ - kt$_{m}$)

S$_{n}$ $\leq$ kt$_{n}$ +h

where h = S$_{m}$ - kt$_{m}$, is a finite quantity

1) If $\sum v_{n}$ is convergent, then the sequence {t$_{n}$} of its partial sums is bounded above, so that there exists a number B such that

t$_{n}$ $\leq$ B, $\forall$ n.

So from 1, we get

S$_{n}$ $\leq$ kB + h, for all n $\geq$ m,

The sequence {S$_{n}$} is bounded above.

$\sum$ u$_{n}$ is convergent.

2)  If $\sum u_{n}$ is divergent, then the sequence {S$_{n}$} of its partial sums is not bounded above, so that if G be any number, however, large, there exists a positive integer m$_{0}$ such that

S$_{n}$ > G $\forall$ n $\geq$ m$_{0}$.

Thus from 1, $\forall$ n $\geq$ max(m,m$_{0}$),

t$_{n}$  $\geq$ $\frac{1}{k}$(G - h), k $\neq$ 0

the sequence {t$_{n}$} is unbounded

$\sum v_{n}$ is divergent.
If $\sum u_{n}$ and $\sum v_{n}$ are two positive term series  and there exists a positive integer m such that

$\frac{u_{n}}{u_{n+1}}$ $\geq$ $\frac{v_{n}}{v_{n+1}}$, $\forall$ n $\geq$ m,

then

1) $\sum u_{n}$ is convergent, if $\sum v_{n}$ is convergent, and

2) $\sum v_{n}$ is divergent, if $\sum u_{n}$ is divergent.


Proof: Let m be a positive integer and S$_{n}$ = u$_{1}$ + u$_{2}$ + ......... + u$_{n}$
and t$_{n}$ = v$_{1}$ + v$_{2}$ + ....... + v$_{n}$

For n $\geq$ m, we have

$\frac{u_{m}}{u_{n}}$ = $\frac{u_{m}}{u_{m+1}}$.$\frac{u_{m+1}}{u_{m+2}}$ ...........  $\frac{u_{n-1}}{u_{n}}$ > $\frac{v_{m}}{v_{m+1}}$.$\frac{v_{m+1}}{v_{m+2}}$....... $\frac{v_{n-1}}{v_{m}}$ = $\frac{v_{m}}{v_{n}}$

u$_{n}$  $\leq$ $\frac{u_{m}}{v_{m}}v_{n}$

Since m is a fixed positive integer, therefore $\frac{u_{m}}{v_{m}}$ is fixed number, say k.Thus $\forall$ n$\geq$ m, we have

u$_{n} \leq kv_{n}$

Hence $\sum u_{n}$ converges if $\sum v_{n}$ converges and $\sum v_{n}$ diverges if $\sum u_{n}$ diverges.
This test is known to be the second type in comparison test.
Given below are few examples on comparision test.
Example 1: Show that the series 1 + $\frac{1}{2!}$ + $\frac{1}{3!}$ + $\frac{1}{4!}$ + ..... is convergent

Solution:
we have
$\frac{1}{2!}$= $\frac{1}{2}$

$\frac{1}{3!}$ < $\frac{1}{2^{2}}$

$\frac{1}{4!}$ < $\frac{1}{2^{3}}$

..................

$\frac{1}{n!}$$\frac{1}{2^{n-1}}$

Therefore 1 + $\frac{1}{2!}$ + $\frac{1}{3!}$ + .......... + $\frac{1}{n!}$ + ....... < 1 + $\frac{1}{2}$ + $\frac{1}{2^{2}}$ + .........

Thus each term of the given series after the second is less than the corresponding term of the convergent geometric series

1 + $\frac{1}{2}$ + $\frac{1}{2^{2}}$ + .........
Thus by test 1, the given series converges.

Example 2: Show that the series

$\frac{1}{(log 2)^{p}}$ + $\frac{1}{(log 3)^{p}}$ + ......... + $\frac{1}{(log n)^{p}}$ + .........

diverges for p > 0.

Solution:
Since $lim _{n-> \infty}$$\frac{(log n)^{p}}{n}$ = 0, therefore

(log n)$^{p}$ < n, $\forall$ n > 1

$\frac{1}{(log n)^{p}}$ > $\frac{1}{n}$ , \$forall$  n > 1

Let us compare the given series with the divergent series

$\frac{1}{2}$  + $\frac{1}{3}$ + $\frac{1}{4}$ + ...

Since each term of the given series is greater than the corresponding term of the divergent series, therefore, the given series diverges.