In mathematics, the limit comparison test (LCT) (in contrast with the related direct comparison test) is a method of testing for the convergence of an infinite series. 

Let sum $a_k$ and sum $b_k$ be a series with positive terms and suppose $a_1$ $\leq$ $b_1$, $a_2$  $\leq$ $b_2$,.....

1) If the bigger series converges, then the smaller series also converges.

2) If the smaller series diverges, then the bigger series also diverges.

Let $\sum u_{n} and \sum v_{n}$ be two series of positive terms, such that,

$\sum v_{n}$ is convergent

and u$_{n}\leq k.v_{n}$, $\forall$ n; except perhaps for the finite number of terms in the beginning where k > 0

Then $\sum u_{n}$ is also convergent.

Proof: Without loss of generality, let us assume that,

u$_{n}\leq k.v_{n} \forall$ n  *

Let s$_{n}$ and S$_{n}$ be the nth partial sums of $\sum u_{n}$ and $\sum v_{n}$ respectively.

s$_{n} = u_{1} + u_{2} + u_{3} + ...... + u_{n}$

and S$_{n} = v_{1} + v_{2} + v_{3} + ...... + v_{n}$

Since $\sum v_{n}$ is convergent, we have,

lim$_{n-> \infty}S_{n}$ = finite and unique, say l,

Now,

s$_{n}= u_{1}+ u_{2}+ ...... + u_{n}$

$\leq kv_{1}+ kv_{2} + .......... + kv_{n}$

=k(v$_{1} + v_{2} + ....... + v_{n}$)

= k. S$_{n}$

s$_{n}$ $\leq k. S_{n}$

lim$_{n->\infty}s_{n}\leq k. lim_{n->\infty}S_{n}$ = k.l

Therefore lim$_{n->\infty}s_{n}$ is finite and unique.

Thus, s$_{n}$ is convergent and hence $\sum u_{n}$ is also convergent.

2) Let $\sum u_{n} and  \sum v_{n}$ be two series of positive terms, such that,
$\sum v_{n}$ is divergent.
 
and u$_{n} \geq kv_{n}$, for all values n, except perhaps for first few finite number of terms. Then

$\sum u_{n}$ is also divergent.

Proof: Without loss of generality, let us assume that

u$_{n} \geq k.v_{n} \forall$ n

Let s$_{n}$ and S$_{n}$ be the nth partial sums of $\sum u_{n}$ and $\sum v_{n}$ respectively.

s$_{n} = u_{1}+ u_{2}+ ...... + u_{n}$

S$_{n}= v_{1} + v_{2} + ...... + v_{n}$

By data, $\sum v_{n}$ is divergent

lim$_{n-> \infty}S_{n} = +\infty$

Now, $s_{n} = u_{1}+ u_{2}+ ...... + u_{n}$

$\geq kv_{1}+ kv_{2} + ....... + kv_{n}$

=k(v$_{1}$ + v$_{2}$ + ....... + v$_{n}$)

= k. S$_{n}$

s$_{n} \geq k. S_{n}$

$lim_{n->\infty}s_{n}\geq k. lim_{n->\infty}S_{n}= + \infty$

Therefore $lim_{n->\infty}s_{n}= + \infty$

Thus, {$s_{n}$}is divergent and hence $\sum u_{n}$ is also divergent.
Let $\sum u_{n}$ and $\sum v_{n}$ be two series of positive terms and $lim_{n->\infty}$ $\frac{u_{n}}{v_{n}}$ be a finite non zero quantity.
 
Then $\sum u_{n}$  and $\sum v_{n}$ both converge or diverge together.

Proof: Let, $lim_{n->\infty}$ $\frac{u_{n}}{v_{n}}$ = l, where l is a finite quantity
Further l > 0 as $\frac{u_{n}}{v_{n}}$ > 0

By the definition of limit, $\forall \in$ > 0 there exists m $\in$ N such that,

|$\frac{u_{n}}{v_{n}}$ - $l$| $< \in, \forall n > m$

- $\in$ < $\frac{u_{n}}{v_{n}}$ - $l< \in, \forall n> m$

$l - \in$ <$ \frac{u_{n}}{v_{n}}$ $< l + \in  \forall n> m$

Rejecting the first m terms of both the series, we have

$l - \in$  < $\frac{u_{n}}{v_{n}}$ $< l + \in  \forall n$.  ---------1

Case 1:

Let $\sum v_{n}$ be convergent.

Then by 1 we have,

$\frac{u_{n}}{v_{n}}$ $< l + \in, \forall n$

$u_{n} < (l + \in). v_{n}, \forall n$.

Therefore by comparision test 1,

$\sum u_{n}$ is also convergent.

case 2: Let $\sum v_{n}$ be divergent

Then again by 1 we have

$\frac{u_{n}}{v_{n}}$ $ > l - \in, \forall$ n

We can choose $\in$ > 0, sufficiently small, such that l - $\in$ > 0.

$u_{n} > (l - \in). v_{n}, \forall n$.

Therefore by comparision test 2, $\sum u_{n}$ is also divergent.

Hence the proof.

Similarly one can also show that, if $\sum u_{n}$ is convergent, then $\sum v_{n}$

is also convergent and if $\sum u_{n}$ is divergent, then $\sum v_{n}$ is also divergent.
Given below are few examples on comparision test.

Example 1: Test the following series for convergence.

$\frac{1}{1.2}$ + $\frac{1}{2.3}$ + $\frac{1}{3.4}$ + .....

Solution: Here $u_{n}$ = $\frac{1}{n.(n + 1)}$

Now consider the series

$\sum v_{n}$ = $\sum$ $\frac{1}{n^{2}}$

Consider, $lim_{n->\infty}$ $\frac{u_{n}}{v_{n}}$ = $lim_{n->\infty}$

$\frac{n^{2}}{n(n+1)}$


= $lim_{n->\infty}$($\frac{1}{(1+\frac{1}{n})}$) = 1 ($\neq$ 0)

By comparision test, $\sum u_{n}$ and $\sum v_{n}$ behave alike

But $\sum v_{n}$ = $\sum$ $\frac{1}{n^{2}}$ is convergent

$\sum u_{n}$ = $\sum$ $\frac{1}{n(n + 1)}$ is convergent


Example 2: Discuss the convergence of the series

$\sqrt{\frac{1}{4}} + \sqrt{\frac{2}{6}} + \sqrt{\frac{3}{8}}$ + .....

Solution: Here $u_{n}$ = $\sqrt{\frac{n}{2(n+1)}}$

u$_{n}$ = $\sqrt{\frac{nth\ term\ of\ 1,\ 2,\ 3\, .....}{nth\ term\ of\ 4,\ 6,\ 8,\ ....}}$

nth term of 4, 6, 8, ........ is 4 + (n - 1)2

= 2(n + 1)

because they form an A.P.

$lim_{n->\infty }u_{n}$ = $lim_{n->\infty}$ $\sqrt{\frac{n}{2(n  + 1)}}$

= $lim_{n->\infty}$ $\sqrt{\frac{1}{2(1 + \frac{1}{n})}}$

= $\frac{1}{\sqrt{2}}$ $(\neq 0)$

Since lim$_{n->\infty}u_{n} \neq 0, \sum u_{n}$ is not convergent.

Therefore $\sum u_{n}$ is divergent.

Example 3 : Test the following series for convergence

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

Here, n$_{n}$ = $\frac{1}{\sqrt{n} + \sqrt{n+1}}$

Consider the series $\sum v_{n}$ = $\sum$ $\frac{1}{\sqrt{n}}$ which is divergent

now, $lim_{n->\infty}$ $\frac{u_{n}}{v_{n}}$ = $lim_{n->\infty}$ $\frac{\sqrt{n}}{\sqrt{n} + \sqrt{n+1}}$

= $lim_{n->\infty}$ $\frac{1}{1 + \sqrt{1 + \frac{1}{n}}}$

= $\frac{1}{1 + \sqrt{1}}$

= $\frac{1}{2}$

Therefore by comparision test $\sum u_{n}$ is also divergent.

Example 4: Test the following series for convergence

$\sum [\sqrt[3]{n + 1} - \sqrt[3]{n}]$

Solution: Here, $u_{n}$ = $\sum [\sqrt[3]{n + 1} - \sqrt[3]{n}]$

 = $(n + 1)^{\frac{1}{3}}$ - $n^{\frac{1}{3}}$

= $n^{\frac{1}{3}} (1 + \frac{1}{n})^{\frac{1}{3}} - n^{\frac{1}{3}}$

= $n^{\frac{1}{3}} [ 1 + \frac{1}{3}. \frac{1}{n} + \frac{\frac{1}{3}(\frac{1}{3}- 1)}{2!}. \frac{1}{n^{2}}+ ...... -1]$

The above expression is obtained using binomial theorem for rational index,

= $\frac{n^{\frac{1}{3}}}{n}[\frac{1}{3} - \frac{1}{9}. \frac{1}{n}.....]$

= $\frac{1}{n^{\frac{2}{3}}}[\frac{1}{3} - \frac{1}{9}. \frac{1}{n}.....]$

Consider $\sum v_{n}$ = $\sum$ $\frac{1}{n^{\frac{2}{3}}}$

which is divergent (as p < 1)

Now, lim$_{n->\infty}$ $\frac{u_{n}}{v_{n}}$ = lim$_{n->\infty}$ $[ \frac{1}{3} - \frac{1}{9}. \frac{1}{n}....]$

= $\frac{1}{3}$
Therefore by comparison test $\sum u_{n}$ is also divergent