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.