Limit comparison test

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.

Statement

Suppose that we have two series $\Sigma_n a_n$ and $\Sigma_n b_n$ with $a_n\geq 0, b_n > 0$ for all $n$.
Then if $\lim_ \frac = c$ with $0 < c < \infty$, then either both series converge or both series diverge.

Proof

Because $\lim_ \frac = c$ we know that for every $\varepsilon > 0$ there is a positive integer $n_0$ such that for all $n \geq n_0$ we have that $\left, \frac - c \ < \varepsilon$, or equivalently

: $- \varepsilon < \frac - c < \varepsilon$

: $c - \varepsilon < \frac < c + \varepsilon$

: $(c - \varepsilon)b_n < a_n < (c + \varepsilon)b_n$
As $c > 0$ we can choose $\varepsilon$ to be sufficiently small such that $c-\varepsilon$ is positive.
So $b_n < \frac a_n$ and by the direct comparison test, if $\sum_n a_n$ converges then so does $\sum_n b_n$.

Similarly $a_n < (c + \varepsilon)b_n$, so if $\sum_n a_n$ diverges, again by the direct comparison test, so does $\sum_n b_n$.

That is, both series converge or both series diverge.

Example

We want to determine if the series $\sum_^ \frac$ converges. For this we compare it with the convergent series $\sum_^ \frac = \frac$

As $\lim_ \frac \frac = 1 > 0$ we have that the original series also converges.

One-sided version

One can state a one-sided comparison test by using limit superior. Let $a_n, b_n \geq 0$ for all $n$. Then if $\limsup_ \frac = c$ with $0 \leq c < \infty$ and $\Sigma_n b_n$ converges, necessarily $\Sigma_n a_n$ converges.

Example

Let $a_n = \frac$ and $b_n = \frac$ for all natural numbers $n$. Now
$\lim_ \frac = \lim_(1-(-1)^n)$ does not exist, so we cannot apply the standard comparison test. However,
$\limsup_ \frac = \limsup_(1-(-1)^n) =2\in [0,\infty)$ and since $\sum_^ \frac$ converges, the one-sided comparison test implies that $\sum_^\frac$ converges.

Converse of the one-sided comparison test

Let $a_n, b_n \geq 0$ for all $n$. If $\Sigma_n a_n$ diverges and $\Sigma_n b_n$ converges, then necessarily
$\limsup_ \frac=\infty$, that is,
$\liminf_ \frac= 0$. The essential content here is that in some sense the numbers $a_n$ are larger than the numbers $b_n$.

Example

Let $f(z)=\sum_^a_nz^n$ be analytic in the unit disc $D = \$ and have image of finite area. By Parseval's formula the area of the image of $f$ is proportional to $\sum_^ n, a_n, ^2$. Moreover,
$\sum_^ 1/n$ diverges. Therefore, by the converse of the comparison test, we have
$\liminf_ \frac= \liminf_ (n, a_n, )^2 = 0$, that is,
$\liminf_ n, a_n, = 0$.

See also

* Convergence tests
* Direct comparison test

References

Further reading

* Rinaldo B. Schinazi: ''From Calculus to Analysis''. Springer, 2011, , pp
50
* Michele Longo and Vincenzo Valori: ''The Comparison Test: Not Just for Nonnegative Series''. Mathematics Magazine, Vol. 79, No. 3 (Jun., 2006), pp. 205–210
JSTOR

* J. Marshall Ash: ''The Limit Comparison Test Needs Positivity''. Mathematics Magazine, Vol. 85, No. 5 (December 2012), pp. 374–375
JSTOR

External links

Pauls Online Notes on Comparison Test

{{Calculus topics

Convergence tests
Articles containing proofs