pg441 ex23-25
In the exercise, use the Limit Comparison test to determine the convergence of the given series; state what series is used for comparison.
\[\sum_{n=1}^{\infty} \frac{1}{n^2-3n+5}\]
Take the dominant term from denominator to form a comparison, I get
\[\frac{1}{n^2}\]
Perform Limit Comparison Test,
\[\begin{align*} & \lim_{n \to \infty} \frac{1/(n^2-3n+5)}{1/n^2} \\ = & \lim_{n \to \infty} \frac{n^2}{n^2-3n+5} \\ = & 1 \end{align*}\]
According to theorem if an/bn = 1, both either converge or diverge and \(\frac{1}{n^2}\) is a p series where p > 1, \(\frac{1}{n^2}\) converges, so \(\frac{1}{n^2-3n+5}\) converges.
\[\sum_{n=1}^{\infty} \frac{1}{4^n-n^2}\]
I am going to use the same methology, and find the comparison
\[\frac{1}{4^n}\]
Then,
\[\begin{align*} & \lim_{n \to \infty} \frac{1/(4^n-n^2)}{1/4^n} \\ = & \lim_{n \to \infty} \frac{4^n}{4^n-n^2} \\ = & 1 \end{align*}\]
Apply the same theorem, and \(\frac{1}{4^n}\) is a geometric series where \(r = \frac{1}{4} \lt 1\). So, \(\frac{1}{4^n}\) converges, \(\frac{1}{4^n-n^2}\) converges.
\[\sum_{n=4}^{\infty} \frac{ln(n)}{n-3}\]
Same procedure as above, the comparison will be
\[\frac{ln(n)}{n}\]
Then,
\[\begin{align*} & \lim_{n \to \infty} \frac{ln(n)/(n-3)}{ln(n)/n} \\ = & \lim_{n \to \infty} \frac{n}{n-3} \\ = & 1 \end{align*}\]
I need to know if \(\frac{ln(n)}{n}\) converges first, so that I will make judgment of whether \(\frac{ln(n)}{n-3}\) converges.
I am going to use Direct Comparison test to see if the comparison converges.
\[\frac{1}{n} < \frac{ln(n)}{n}\] for n >= 3
\(\frac{1}{n}\) is a harmony series which diverges, so \(\frac{ln(n)}{n}\) diverges. As a result, \(\frac{ln(n)}{n-3}\) diverges as well.