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Limit Comparison Test

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What is the Limit Comparison Test ???

Suppose that $$\sum a_{n}$$ and $$\sum b_{n}$$ are series with positive terms.
If $$\lim\limits_{n\rightarrow \infty}\frac{a_{n}}{b_{n}}=\rho , 0<\rho<\infty $$,
then both converge or both diverge .
If $$ \lim\limits_{n\rightarrow \infty}\frac{a_{n}}{b_{n}}=0 \text{, and } \sum b_{n}$$ converges
then $$\sum a_{n}$$ converges
If $$ \lim\limits_{n\rightarrow \infty}\frac{a_{n}}{b_{n}}=\infty \text{, and } \sum b_{n} $$ diverges
Then $$\sum a_{n}$$ diverges.

Notes

  • Basically, the test is performed by taking a ratio of the general terms of two series with positive terms. If the limit of the ratio is a positive finite number, then the two series have the same convergence properties.
  • We can find a suitable comparison series $$\sum b_{n}$$ by keeping only the highest powers in the numerator and denominator of the fraction.
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What are the benefits of this test?

Examples of the Limit Comparison Test

$ \text{Example 1)} \sum\limits_{n=1}^{\infty}\frac{\sqrt{n}+2}{n^2+3n+1}$
Find Comparison Series
Taking the highest power of n in the numerator and denominator we get $$\frac{n^{\frac{1}{2}}}{n^2}=\frac{1}{n^{\frac{3}{2}}}$$
Find The limit of the Ratio
$$\rho=\lim\limits_{n\rightarrow \infty} \Bigg(\frac{\frac{\sqrt{n}+2}{n^2+3n+1}}{\frac{1}{n^{\frac{3}{2}}}}\Bigg)=\lim\limits_{n\rightarrow \infty}\big(\frac{n^2+2n^{\frac{3}{2}}}{n^2+3n+1}\big)=1$$
Conclusion:
Since $$\rho=1 (0<\rho <\infty)$$, both series behave the same way. The P-series $$\sum\limits_{n=1}^{\infty}\frac{1}{n^{\frac{3}{2}}} $$ converges which means the series $$\sum\limits_{n=1}^{\infty}\frac{\sqrt{n}+2}{n^2+3n+1}$$ also converges.
$\text{Example 2)} \sum\limits_{n=1}^{\infty}\frac{1}{\sqrt{5n-3}}$
Find Comparison Series
Taking the highest power of n in the numerator and denominator $$\frac{1}{\sqrt{n}}$$
Find The limit of the Ratio
$$\rho=\lim\limits_{n\rightarrow \infty} \big(\frac{\frac{1}{\sqrt{5n-3}}}{\frac{1}{\sqrt{n}}}\big)=\frac{\sqrt{n}}{\sqrt{5n-3}}=\frac{1}{\sqrt{5}}$$
Conclusion:
$$\rho=\frac{1}{\sqrt{5}} (0<\rho <\infty)$$, both series behave the same way. The P-series $$\sum\limits_{n=1}^{\infty}\frac{1}{\sqrt{n}}$$ diverges which means that $$\sum\limits_{n=1}^{\infty}\frac{1}{\sqrt{5n-3}}$$ also diverges.