If the limit of the ratio between the terms of the two series approaches 1, then the original series will have the same convergence behavior as the comparison series.

For example, consider the series:

n=1 (2^(-n)) / n!

To apply the Limit Comparison Test, we need to find a comparison series that has already been proven convergent or divergent. A good choice for this is the p-series with p = 2:

n=1 1/n^2

The limit of the ratio between these two terms as n approaches infinity is:

lim (n ) [(2^(-n)) / n!] / [1/n^2]

This simplifies to:

lim (n ) [n^2 * 2^(-n)] / [n!]

Using Stirling’s approximation, we can approximate the factorial term as follows:

n! ~ sqrt(2πn) * n^(n+1/2) * e^-n

Substituting this into our limit gives us:

lim (n ) [n^2 * 2^(-n)] / [sqrt(2πn) * n^(n+1/2) * e^-n]

This simplifies to:

lim (n ) [2^(-n)] / [e^-n * sqrt(2πn) * n^{-3/2}]

Taking the limit, we get:

lim (n ) [2^(-n)] / [e^-n * sqrt(2πn) * n^{-3/2}] = e^2 / sqrt(2π) < 1 Since this ratio is less than 1 and approaches a constant value, we can conclude that the original series converges by comparison with the p-series. This result agrees with our intuition since the terms in the original series are decreasing exponentially while the factorial term grows more slowly than n^2.