# limit comparison test

The Limit Comparison Test
If the limit of a[n]/b[n] is positive, then the sum of a[n] converges if and only if the sum of b[n] converges. If the limit of a[n]/b[n] is zero, and the sum of b[n] converges, then the sum of a[n] also converges.

this section we will discuss using the Comparison Test and Limit Comparison Tests to determine if an infinite series converges or diverges.
In mathematics, the limit comparison test (LCT) is a method of testing for the convergence of an infinite series. Contents. 1 Statement; 2 Proof; 3 Example .

## Statement

Suppose that we have two series ${\displaystyle \Sigma _{n}a_{n}}$ and ${\displaystyle \Sigma _{n}b_{n}}$ with ${\displaystyle a_{n}\geq 0,b_{n}>0}$ for all ${\displaystyle n}$.
Then if ${\displaystyle \lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}=c}$ with ${\displaystyle 0, then either both series converge or both series diverge.[1]

## Proof

Because ${\displaystyle \lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}=c}$ we know that for all ${\displaystyle \varepsilon >0}$ there is a positive integer ${\displaystyle n_{0}}$ such that for all ${\displaystyle n\geq n_{0}}$ we have that ${\displaystyle \left|{\frac {a_{n}}{b_{n}}}-c\right|<\varepsilon }$, or equivalently
${\displaystyle -\varepsilon <{\frac {a_{n}}{b_{n}}}-c<\varepsilon }$
${\displaystyle c-\varepsilon <{\frac {a_{n}}{b_{n}}}
${\displaystyle (c-\varepsilon )b_{n}
As ${\displaystyle c>0}$ we can choose ${\displaystyle \varepsilon }$ to be sufficiently small such that ${\displaystyle c-\varepsilon }$ is positive. So ${\displaystyle b_{n}<{\frac {1}{c-\varepsilon }}a_{n}}$ and by the direct comparison test, if ${\displaystyle \sum _{n}a_{n}}$ converges then so does ${\displaystyle \sum _{n}b_{n}}$.
Similarly ${\displaystyle a_{n}<(c+\varepsilon )b_{n}}$, so if ${\displaystyle \sum _{n}a_{n}}$ diverges, again by the direct comparison test, so does ${\displaystyle \sum _{n}b_{n}}$.
That is, both series converge or both series diverge.

## Example

We want to determine if the series ${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}+2n}}}$ converges. For this we compare with the convergent series ${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}={\frac {\pi ^{2}}{6}}}$.
As ${\displaystyle \lim _{n\to \infty }{\frac {1}{n^{2}+2n}}{\frac {n^{2}}{1}}=1>0}$ we have that the original series also converges.

## One-sided version

One can state a one-sided comparison test by using limit superior. Let ${\displaystyle a_{n},b_{n}\geq 0}$ for all ${\displaystyle n}$. Then if ${\displaystyle \limsup _{n\to \infty }{\frac {a_{n}}{b_{n}}}=c}$ with ${\displaystyle 0\leq c<\infty }$ and ${\displaystyle \Sigma _{n}b_{n}}$ converges, necessarily ${\displaystyle \Sigma _{n}a_{n}}$ converges.

## Example

Let ${\displaystyle a_{n}={\frac {1-(-1)^{n}}{n^{2}}}}$ and ${\displaystyle b_{n}={\frac {1}{n^{2}}}}$ for all natural numbers ${\displaystyle n}$. Now ${\displaystyle \lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}=\lim _{n\to \infty }(1-(-1)^{n})}$ does not exist, so we cannot apply the standard comparison test. However, ${\displaystyle \limsup _{n\to \infty }{\frac {a_{n}}{b_{n}}}=\limsup _{n\to \infty }(1-(-1)^{n})=2\in [0,\infty )}$ and since ${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}}$ converges, the one-sided comparison test implies that ${\displaystyle \sum _{n=1}^{\infty }{\frac {1-(-1)^{n}}{n^{2}}}}$ converges.

## Converse of the one-sided comparison test

Let ${\displaystyle a_{n},b_{n}\geq 0}$ for all ${\displaystyle n}$. If ${\displaystyle \Sigma _{n}a_{n}}$ diverges and ${\displaystyle \Sigma _{n}b_{n}}$ converges, then necessarily ${\displaystyle \limsup _{n\to \infty }{\frac {a_{n}}{b_{n}}}=\infty }$, that is, ${\displaystyle \liminf _{n\to \infty }{\frac {b_{n}}{a_{n}}}=0}$. The essential content here is that in some sense the numbers ${\displaystyle a_{n}}$ are larger than the numbers ${\displaystyle b_{n}}$.

## Example

Let ${\displaystyle f(z)=\sum _{n=0}^{\infty }a_{n}z^{n}}$ be analytic in the unit disc ${\displaystyle D=\{z\in \mathbb {C} :|z|<1\}}$ and have image of finite area. By Parseval's formula the area of the image of ${\displaystyle f}$ is ${\displaystyle \sum _{n=1}^{\infty }n|a_{n}|^{2}}$. Moreover, ${\displaystyle \sum _{n=1}^{\infty }1/n}$ diverges. Therefore, by the converse of the comparison test, we have ${\displaystyle \liminf _{n\to \infty }{\frac {n|a_{n}|^{2}}{1/n}}=\liminf _{n\to \infty }(n|a_{n}|)^{2}=0}$, that is, ${\displaystyle \liminf _{n\to \infty }n|a_{n}|=0}$.

The limit comparison test shows that the original series is divergent. The limit comparison test does not apply because the limit in question does not exist. The ever useful Limit Comparison Test will save the day! using the comparison tests to determine convergence of an infinite series. Comparison Test Example 1 Example 2 Example 3 Example 4 Example 5 Example 6 Limit Comparison Test Example Example Example Example Example.

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