Infinite Series Practice Problem 2437 

\(\displaystyle{\sum_{n=0}^{\infty}{\frac{1}{3^n+n}}}\) 
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. Also, 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 \(3^n\) which is in the denominator added to another term. This could be a problem to integrate. So we set aside the use of the integral test for the moment.
What We Are Left With 

4. Now we are left with the ratio test, the comparison tests and the root test. By now, you know that all the tests in that list, except for the direct comparison test, involve limits. When we first tried this problem, we went straight for the ratio test but once we set up the limit, we noticed that we had to use L'Hôpital's Rule and take the derivative of \(3^x\). Although this is quite doable, we don't come across this very often and we might be a bit rusty on how to do it. So, rather than do that, we jumped to the only test that doesn't involve a limit, the direct comparison test.
Applying The Tests 

The ratio test requires us to set up the limit \(\displaystyle{ \lim_{n \to \infty}{ \left \frac{a_{n+1}}{a_n} \right } }\).
\(\displaystyle{ a_n = \frac{1}{3^n+n} ~~~ \to ~~~ a_{n+1} = \frac{1}{3^{n+1}+n+1} }\)
\(
\begin{array}{rclr}
\displaystyle{\lim_{n \to \infty}{\left \frac{a_{n+1}}{a_n} \right}}
& = & \displaystyle{\lim_{n \to \infty}{\left \frac{1}{3^{n+1}+n+1} \frac{3^n+n}{1} \right }} \\
& = & \displaystyle{\lim_{n \to \infty}{ \left \frac{3^n + n}{3^{n+1}+n+1} \right }} \\
& = & \displaystyle{\lim_{n \to \infty}{ \left \frac{3^n + n}{3^{n+1}+n+1} \frac{1/3^n}{1/3^n} \right }} \\
& = & \displaystyle{\lim_{n \to \infty}{ \left \frac{1 + n/3^n}{3+n/3^n+1/3^n} \right }} & ~~~ (1)
\end{array}
\)
At this point, we can take the limit of each term but we will have difficulty with the term \(n/3^n\) since the limit is \(\infty / \infty\) which is indeterminate. So, we will use L'Hôpital's Rule to evaluate it.
\(\displaystyle{
\lim_{x \to \infty}{ \frac{x}{3^x} } = \lim_{x \to \infty}{ \frac{1}{(\ln 3)3^x} } = 0
}\)
So now we can finish the limit in equation \((1)\) above
\(\displaystyle{
\lim_{n \to \infty}{\left \frac{a_{n+1}}{a_n} \right} = \lim_{n \to \infty}{ \left \frac{1 + n/3^n}{3+n/3^n+1/3^n} \right } = \left \frac{1+0}{3+0+0} \right = \frac{1}{3}
}\)
Conclusion
Since
\(\displaystyle{
\lim_{n \to \infty}{\left \frac{a_{n+1}}{a_n} \right} = \frac{1}{3}
}\)
Since the limit \(1/3 < 1\), the series \(\sum{a_n}\) converges by the ratio test.
Notes:
1. In order to use L'Hôpital's Rule, we had to calculate the derivative of \(3^x\). There are two ways to remember this. The first is to use what you know about exponentials to convert \(3^x\) to \(e^{x\ln3}\) and then taking the derivative of the exponential. The other way is to just know that \((k^x)' = \ln(k)k^x\) where \(k\) is a constant.
2. You may have noticed that we changed the variable from \(n\) to \(x\) when we used L'Hôpital's Rule. This is a subtle but important point. Derivatives are defined only on continuous functions. You can find more details on the L'Hôpital's Rule page.
\(\displaystyle{ a_n = \frac{1}{3^n+n} }\)
We need to choose a test series for comparison. Since \(3^n >> n\) for large \(n\) let's compare this series to \(\displaystyle{ t_n = \frac{1}{3^n}}\), which is a convergent pseries.
The limit comparison test requires us to find the limit \(\displaystyle{ \lim_{n \to \infty}{\frac{a_n}{t_n}} }\) or
\(\displaystyle{ \lim_{n \to \infty}{\frac{t_n}{a_n}} }\). Looking at these terms, it looks like the second option is easier.
\(
\begin{array}{rcl}
\displaystyle{\lim_{n \to \infty}{\frac{t_n}{a_n}}}
& = & \displaystyle{\lim_{n \to \infty}{\frac{1}{3^n} \frac{3^n+n}{1}}} \\
& = & \displaystyle{\lim_{n \to \infty}{\frac{3^n+n}{3^n}}} \\
& = & \displaystyle{\lim_{n \to \infty}{ \left[ 1 + \frac{n}{3^n} \right]} = 1 + 0 = 1}
\end{array}
\)
Conclusion
Since the limit \(\displaystyle{ \lim_{n \to \infty}{\frac{t_n}{a_n}} = 1 }\) which is positive and finite, the series converges by the limit comparison test.
Note
The limit \(\displaystyle{ \lim_{n \to \infty}{ \left[ \frac{n}{3^n} \right]} = 0 }\) is discussed in detail in the previous panel.
\(\displaystyle{ a_n = \frac{1}{3^n+n} }\)
Since \(3^n >> n\) for large \(n\), let's compare this series with the test series \(\sum{t_n}\) where \(\displaystyle{t_n = \frac{1}{3^n} }\).
Since \(t_n\) is a pseries with \(p=3>1\), the test series converges. So the direct comparison test requires us to set up the inequality as \(a_n < t_n\).
\(
\begin{array}{rcl}
a_n & < & t_n \\
\displaystyle{\frac{1}{3^n+n}} & < & \displaystyle{\frac{1}{3^n}} \\
3^n & < & 3^n + n \\
0 & < & n
\end{array}
\)
Conclusion
This last inequality is true for \(N=1\) and \(n>N\). So the series \(\sum{a_n}\) converges by the direct comparison test.
\(\displaystyle{ a_n = \frac{1}{3^n+n} }\)
The root test requires us to evaluate the limit \(\displaystyle{ \lim_{n \to \infty}{ \sqrt[n]{a_n} } }\). We will drop the absolute value signs since \(a_n > 0\) for all \(n\).
\(\displaystyle{
\lim_{n \to \infty}{ \sqrt[n]{a_n} } = \lim_{n \to \infty}{ \sqrt[n]{ \frac{1}{3^n+n} } }
= \lim_{n \to \infty}{ \frac{1}{ \sqrt[n]{ 3^n+n } } }
}\)
We need to evaluate the limit of the denominator, as follows
\(
\begin{array}{rcl}
y & = & \displaystyle{\lim_{n \to \infty}{ \sqrt[n]{ 3^n+n } }} \\
\ln(y) & = & \displaystyle{\lim_{n \to \infty}{ (1/n) \ln( 3^n+n ) }} \\
& = & \displaystyle{\lim_{n \to \infty}{ \frac{\ln( 3^n+n ) }{n} }} ~~~~~ (1)
\end{array}
\)
The last limit above is \(\infty / \infty \) which is indeterminate in the correct form to use L'Hôpital's Rule.
\(
\begin{array}{rcl}
\displaystyle{\lim_{x \to \infty}{ \frac{\ln( 3^x+x ) }{x} }}
& = & \displaystyle{\lim_{x \to \infty}{ \frac{1}{3^x+x} \frac{ (\ln3)3^x +1 }{1} }} \\
& = & \displaystyle{\lim_{x \to \infty}{ \frac{(\ln3)3^x +1}{3^x+x} \frac{ 1/3^x }{1/3^x} }} \\
& = & \displaystyle{\lim_{x \to \infty}{ \frac{\ln(3) + 1/3^x}{1+x/3^x} } = \frac{\ln(3)+0}{1+0} = \ln(3)} \\
\end{array}
\)
Finishing equation (1) above, we have
\(\displaystyle{
\ln(y) = \lim_{n \to \infty}{ \frac{\ln( 3^n+n ) }{n} } = \ln(3) ~~~ \to ~~~ y = 3
}\)
And so the limit is
\(\displaystyle{
\lim_{n \to \infty}{ \sqrt[n]{a_n} } = 1/y = 1/3 < 1
}\)
Conclusion
The series converges by the root test.
Note
The limit \(\displaystyle{ \lim_{n \to \infty}{ \left[ \frac{n}{3^n} \right]} = 0 }\) is discussed in detail in the first panel.
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 

all terms are positive, so this is not an alternating series  
no 

not a telescoping series  

yes 
 

yes 
 

yes 
best test  
no 

not integrable  

yes 

Final Answer 

The series \(\displaystyle{\sum_{n=0}^{\infty}{\frac{1}{3^n+n}}}\) converges.
Notes 

The ratio, limit comparison and root tests required us to evaluate a limit involving \(3^n\) and we needed to use L'Hopitals Rule, which in turn required us to take the derivative of \(3^n\). The direct comparison test did not require this. We had to prove only a simple inequality.
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L'Hopitals Rule 
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