Infinite Series Practice Problem 2438 

\(\displaystyle{\sum_{n=1}^{\infty}{\frac{n}{n^2\cos^2(n)}}}\) 
Determine the convergence or divergence of the series. 
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Choosing Which Tests To Try 

Rejecting the Obvious 

1. We can tell right away that this is not an alternating series, telescoping series or one of the special series.
2. We can tell that the limit of \(a_n\) is zero. So the divergence test will not help us.
3. The form of \(a_n\) contains \(\cos^2(n)\) in the denominator and a polynomial in the denominator. This combination is a problem to integrate.
What We Are Left With 

4. We are left with the ratio test, the comparison tests and the root test. The root test will give us quite a complicated term in the denominator. So we will not try that one unless nothing else works.
Applying The Tests 

\(\displaystyle{ a_n = \frac{n}{n^2\cos^2(n)} }\) and \(\displaystyle{ a_{n+1} = \frac{n+1}{(n+1)^2\cos^2(n+1)} }\)
\(\displaystyle{
\begin{array}{rcl}
\displaystyle{\lim_{n \to \infty}{ \left \frac{a_{n+1}}{a_n} \right }} & = &
\displaystyle{\lim_{n \to \infty}{ \left \frac{n+1}{(n+1)^2\cos^2(n+1)} \frac{n^2\cos^2(n)}{n} \right }} \\
& = & \displaystyle{\lim_{n \to \infty}{ \left \frac{n+1}{n} \frac{n^2\cos^2(n)}{(n+1)^2\cos^2(n+1)} \right }} \\
& = & \displaystyle{\lim_{n \to \infty}{ \left \left[ 1 + \frac{1}{n} \right] \frac{n^2\cos^2(n)}{(n+1)^2\cos^2(n+1)}\frac{1/n^2}{1/n^2} \right }} \\
& = & \displaystyle{\lim_{n \to \infty}{ \left \left[ 1 + \frac{1}{n} \right] \frac{1\cos^2(n)/n^2}{(1+1/n)^2\cos^2(n+1)/n^2} \right } }
\end{array}
}\)
Now let's look at each piece.
\(\displaystyle{ \lim_{n \to \infty}{ 1/n } = 0 }\)
Using the pinching theorem, we know that \(\displaystyle{ \lim_{n \to \infty}{ \frac{\cos^2(n)}{n^2} } =0 }\) and
\(\displaystyle{ \lim_{n \to \infty}{ \frac{\cos^2(n+1)}{n^2} } =0 }\). The details are in the section below.
So now we have
\(\displaystyle{
\lim_{n \to \infty}{ \left \left[ 1 + \frac{1}{n} \right] \frac{1\cos^2(n)/n^2}{(1+1/n)^2\cos^2(n+1)/n^2} \right }
= (1+0)\frac{10}{10} = 1
}\)
Conclusion
Since the limit \(\displaystyle{ \lim_{n \to \infty}{ \left \frac{a_{n+1}}{a_n} \right } = 1}\) the ratio test is inconclusive.
\(\displaystyle{ a_n = \frac{n}{n^2\cos^2(n)} }\)
For large \(n\) the \(n^2\) term dominates in the denominator since \(\cos^2(n) \leq 1 \). In the numerator, we have \(n\). So let's compare this series with the test series \(\sum{t_n}\) where \(\displaystyle{t_n = \frac{1}{n} }\).
Since \(t_n\) is a pseries with \(p=1\), the test series diverges. The limit comparison test requires us to set up the limit
\(\displaystyle{ \lim_{n \to \infty}{\frac{a_n}{t_n}} }\).
\(
\begin{array}{rcl}
\displaystyle{\lim_{n \to \infty}{\frac{a_n}{t_n}}}
& = & \displaystyle{\lim_{n \to \infty}{\frac{n}{n^2\cos^2(n)} \frac{n}{1} }} \\
& = & \displaystyle{\lim_{n \to \infty}{\frac{n^2}{n^2\cos^2(n)} \frac{1/n^2}{1/n^2} }} \\
& = & \displaystyle{\lim_{n \to \infty}{\frac{1}{1\cos^2(n)/n^2} }} \\
& = & \displaystyle{\frac{1}{10}} = 1
\end{array}
\)
Since the limit is finite and positive and the test series diverges, the series \(\sum{a_n}\) also diverges.
Conclusion
The series \(\sum{a_n}\) diverges by the limit comparison test.
We evaluated the limit \(\displaystyle{ \lim_{n \to \infty}{\frac{\cos^2(n)}{n^2}} }\) using the pinching theorem. The details can be found in the section below.
\(\displaystyle{ a_n = \frac{n}{n^2\cos^2(n)} }\)
For large \(n\) the \(n^2\) term dominates in the denominator since \(\cos^2(n) \leq 1 \). In the numerator, we have \(n\). So let's compare this series with the test series \(\sum{t_n}\) where \(\displaystyle{t_n = \frac{1}{n} }\).
Since \(t_n\) is a pseries with \(p=1\), the test series diverges. So the direct comparison test requires us to set up the inequality as \(t_n \leq a_n\).
\(
\begin{array}{rcl}
t_n & \leq & a_n \\
\displaystyle{\frac{1}{n}} & \leq & \displaystyle{\frac{n}{n^2  \cos^2(n)}} \\
n^2  \cos^2(n) & \leq & n^2 \\
\cos^2(n) & \leq & 0 \\
\cos^2(n) & \geq & 0
\end{array}
\)
The last inequality is always true since a squared term is always greater than or equal to zero. So the series diverges.
Conclusion
The series \(\sum{a_n}\) diverges by the direct comparison test.
Evaluating Limit With The Pinching Theorem 

\(\begin{array}{rcccl}
0 & \leq & \displaystyle{\frac{\cos^2(n)}{n^2}}
& \leq & \displaystyle{\frac{1}{n^2}} \\
\displaystyle{\lim_{n \to \infty}{0}}
& \leq & \displaystyle{\lim_{n \to \infty}{\frac{\cos^2(n)}{n^2} } }
& \leq & \displaystyle{\lim_{n \to \infty}{\frac{1}{n^2}}} \\
0 & \leq & \displaystyle{\lim_{n \to \infty}{\frac{\cos^2(n)}{n^2} }} & \leq & 0
\end{array}\)
By the pinching theorem, \(\displaystyle{ \lim_{n \to \infty}{\frac{\cos^2(n)}{n^2} } = 0 }\)
Using the same logic, we can say that \(\displaystyle{ \lim_{n \to \infty}{\frac{\cos^2(n+1)}{n^2} } = 0 }\)
Test Summary List and Answer 

test/series  works?  notes  

no 

limit is zero, so test is inconclusive  
no 

not a pseries  
no 

not a geometric series  
no 

not an alternating series  
no 

not a telescoping series  
no 

inconclusive  

yes 
 

yes 
best test  
no 

not integrable  
no 

too complicated 
Final Answer 

The series \(\displaystyle{\sum_{n=1}^{\infty}{\frac{n}{n^2\cos^2(n)}}}\) diverges.
Notes 

The direct comparison test is probably the best test to use here since we did not have to evaluate a limit that involved using the pinching theorem. However, the limit comparison test worked well too.
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