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 p-series.
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 p-series 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.