Section 10.7 : Comparison Test/Limit Comparison Test
8. Determine if the following series converges or diverges.
\[\sum\limits_{n = 3}^\infty {\frac{{{{\bf{e}}^{ - n}}}}{{{n^2} + 2n}}} \]Show All Steps Hide All Steps
Start SolutionFirst, the series terms are,
\[{a_n} = \frac{{{{\bf{e}}^{ - n}}}}{{{n^2} + 2n}}\]and it should pretty obvious that for the range of \(n\) we have in this series that they are positive and so we know that we can attempt the Comparison Test for this series.
It is very important to always check the conditions for a particular series test prior to actually using the test. One of the biggest mistakes that many students make with the series test is using a test on a series that don’t meet the conditions for the test and getting the wrong answer because of that!
In this case let’s first notice the exponential in the numerator will go to zero as \(n\) goes to infinity. Let’s also notice that the denominator is just a polynomial. In cases like this the exponential is going to go to zero so fast that behavior of the denominator will not matter at all and in all probability the series in the problem statement will probably converge as well.
So, because we’re guessing that the series converge we’ll need to find a series with larger terms that we know, or can prove, converge.
Note as well that we’ll also need to prove that the new series terms really are larger than the terms from the series in the problem statement. We can’t just “hope” that the will be larger.
In this case, because the terms in the problem statement series are a rational expression, we know that we can make the series terms larger by either making the numerator larger or the denominator smaller.
In this case we can work with both the numerator and the denominator. Let’s start with the numerator. We can use some quick Calculus I to prove that \({{\bf{e}}^{ - n}}\) is a decreasing function and so,
\[{{\bf{e}}^{ - n}} < {{\bf{e}}^{ - 3}} < 1\]Using this we can make the numerator larger to get the following relationship,
\[\frac{{{{\bf{e}}^{ - n}}}}{{{n^2} + 2n}} < \frac{1}{{{n^2} + 2n}}\]Now, in the denominator it should be fairly clear that,
\[{n^2} + 2n > {n^2}\]Using this we can make the denominator smaller (and hence make the rational expression larger) to get,
\[\frac{{{{\bf{e}}^{ - n}}}}{{{n^2} + 2n}} < \frac{1}{{{n^2} + 2n}} < \frac{1}{{{n^2}}}\] Show Step 4Now, the series,
\[\sum\limits_{n = 3}^\infty {\frac{1}{{{n^2}}}} \]is a convergent series (\(p\)-series Test with \(p = 2 > 1\)) and we’ve also shown that the series terms in this series are larger than the series terms from the series in the problem statement.
Therefore, by the Comparison Test the series given in the problem statement must also converge.