Infinite Series Practice Problem 2440 

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2+4} } }\) 
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 other special series. So we will not be able to determine what the series converges to.
2. We can tell that the limit of \(a_n\) is zero. So the divergence test will not help us.
What We Are Left With 

3. We are left with the ratio test, the comparison tests, the integral test and the root test.
Applying The Tests 

Compare to \(\sum{1/n^2}\), which is a convergent pseries with \(p = 2\)
\(\displaystyle{a_n=\frac{1}{n^2+4}}\), \(\displaystyle{t_n=\frac{1}{n^2}}\)
\(\begin{array}{rcl}
\displaystyle{ \lim_{n\to\infty}{ \frac{a_n}{t_n} } }
& = & \displaystyle{ \lim_{n\to\infty}{ \left[ \frac{1}{n^2+4} \cdot \frac{n^2}{1} \right] } } \\
& = & \displaystyle{ \lim_{n\to\infty}{ \left[ \frac{n^2}{n^2+4} \frac{1/n^2}{1/n^2} \right] } } \\
& = & \displaystyle{ \lim_{n\to\infty}{ \frac{1}{1+4/n^2} } = 1 }
\end{array}\)
Since \(\displaystyle{ 0 \leq \lim_{n\to\infty}{ \frac{a_n}{t_n} } = 1 \lt \infty }\) and \(\sum{t_n}\) converges, the series \(\displaystyle{\sum{\frac{1}{n^2+4}}}\) also converges by the limit comparison test.
Compare to \(\sum{1/n^2}\), which is a convergent pseries with \(p = 2\)
Since the test series \(\sum{1/n^2}\) is convergent, we assume convergence. This tells us we need to show that the inequality \(0\lt a_n \leq t_n\) holds.
\(\displaystyle{a_n=\frac{1}{n^2+4}}\), \(\displaystyle{t_n=\frac{1}{n^2}}\)
It is obvious that \(a_n \gt 0\), so that part of the inequality holds. Now we need to show that \(a_n \leq t_n\).
\(\begin{array}{rcl}
a_n & \leq & t_n \\
\displaystyle{\frac{1}{n^2+4}} & \leq & \displaystyle{\frac{1}{n^2}} \\
n^2 & \leq & n^2+4 \\
0 & \leq & 4
\end{array}\)
The last inequality holds for all \(n\). Therefore since \(0\lt a_n \leq t_n\) holds and \(\sum{t_n}\) converges, the series \(\displaystyle{\sum{\frac{1}{n^2+4}}}\) also converges by the direct comparison test.
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 
best test(?)  

yes 
 

yes 
 
no 

inconclusive 
Final Answer 

The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2+4} } }\) converges.
Notes 

1. Which test is the 'best' to use for this solution is based on whichever test you are most comfortable using. The root test is the longest and takes the most work to show that the denominator goes to \(1\). We suggest the limit comparison test or the direct comparison test.
2. Since \(\sum{\lefta_n\right}\) converges, the original series converges absolutely by the absolute convergence theorem. This is true by default for all series with positive terms.
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