The comparison test lets us compare an unknown series to a simpler, known series. But it requires us to come up with inequalities that can be difficult to figure out. Worse, sometimes it just doesn’t quite work. For example, suppose we’re looking at the series
![Rendered by QuickLaTeX.com \[ \sum_{n=2}^{\infty}\frac{1}{n^2-1}. \]](http://thecalculusblog.com/wp-content/ql-cache/quicklatex.com-4be60e9befc0d16bf3185bb9fd43c919_l3.png)
This series looks almost like a p-series with
, so we expect it to converge. However, if we try to use the comparison test to show that, we end up comparing
with
. We have
![Rendered by QuickLaTeX.com \[ n^2-1 < n^2, \]](http://thecalculusblog.com/wp-content/ql-cache/quicklatex.com-8512fed12bf2e3a02489f81cbc9399b0_l3.png)
so
![Rendered by QuickLaTeX.com \[ \frac{1}{n^2-1} > \frac{1}{n^2}. \]](http://thecalculusblog.com/wp-content/ql-cache/quicklatex.com-39adad75383a9263971753caae260742_l3.png)
But that’s not the inequality we want! Using this inequality we can only show that our series has terms that are larger than the corresponding terms of a converging series. That doesn’t say anything about whether or not our series converges. So how can we proceed?
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