I'm working on the section involving the Integral Test for Convergence in my calculus II class right now, and I've run into a seeming conflict between the definition of the Integral Test, and the solutions to some of the homework exercises as given by both my professor and the textbook.

According to the definition, my research, and my understanding of the integral test, the integral test can only be used for the series $a_n$ where $ a_n = f(n) $ and $ f(x) $is positive, continuous and decreasing for all $ x \ge N $, where $N$ is the index of $n$. However, there are several problems where $f(x)$ is only decreasing if we add a condition, such as $f'(x) < 0 \Leftarrow\Rightarrow x > 3$, where $N \lt 3$. It seems to me that the Integral Test cannot be used to determine convergence of a series when the function is only decreasing when $ x \gt k $ and $ k \lt N $, yet the book and my professor apply the test anyway.

For example, with the series:

$$ \sum_{n=1}^\infty = \frac{n}{(4n+5)^\frac{3}{2}} $$

If we let $a_n = f(n)$, then for $f(x)$:

  • $f(x) \gt 0 $ for all $x$ in the domain
  • $f(x)$ is continuous for all $x \gt -\frac{5}{4} $

But, $f'(x) \lt 0 $ only when $ x > \frac{5}{2} $, as shown when testing the critical points with the derivative:

$$ f'(x) = \frac{5-2x}{(4x+5)^\frac{5}{2}}$$

The professor notes this in her solution, but instead of ending with that and writing, "The Integral Test cannot be applied because $f(x)$ fails to satisfy the required conditions," she applies the test using the original index for $n$:

$$ \int_1^\infty \frac{x}{(4x+5)^\frac{3}{2}} \rightarrow \infty \Rightarrow a_n \text{ Diverges} $$

The textbook reaches the same conclusion. Also, the problems in question are listed under a section where the instructions state, "Confirm that the Integral test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series," implying the test can be used on the subsequent exercises.

Is there a reason the Integral Test for Convergence can be used to test for convergence in these problems, where $N \lt k $ and $f'(x) < 0 \Leftarrow\Rightarrow x \gt k $? Am I missing something, or are the book and my professor wrongly using the Integral Test for these series?

Other exercises with same result:

$$\bullet \sum_{n=1}^\infty \frac{\ln n}{n^2} $$

$$\bullet \frac{\ln 2}{2} + \frac{\ln 3}{3} + \frac{\ln 4}{4} + \frac{\ln 5}{5} + \frac{\ln 6}{6} $$

  • $\begingroup$ One way to look at what your book and your professor are doing is that they're using the fact that $\sum_{n=1}^{\infty}a_n$ converges iff $\sum_{n=k}^{\infty}a_n$ converges, for any $k$. $\endgroup$
    – user84413
    Commented Jul 14, 2014 at 0:20
  • $\begingroup$ I think the idea is more that we don't really care what happens for the first 10, 100, or 1000 terms. We care about what eventually happens. We can add up a finite amount of things no problem; it's the tail end behavior that we really care about. $\endgroup$ Commented Jul 14, 2014 at 0:20
  • $\begingroup$ For convergence, doesn't matter. For estimates it does. $\endgroup$ Commented Jul 14, 2014 at 0:22
  • $\begingroup$ Thank you for your responses; I agree that the behavior is properly demonstrated with the test, and I also understand our concern being only with the end behavior of the function, but I don't think it explains why the test is defined as having to be decreasing for all $x \ge N$, if we only care about the last few terms. Wouldn't it be better to say "for most of"? At any rate, if my definition of the test is correct, doesn't this series still violate the conditions, regardless of the validity of the result? $\endgroup$
    – MrCMedlin
    Commented Jul 14, 2014 at 0:34

2 Answers 2


The point, which isn't made often enough (in my opinion) in classes on the subject, is that the convergence of a series is a limit process. In this case, what that means is that the question of convergence is completely determined by the behavior of the series for say $n > N$ for any fixed, finite $N$. If I take a convergent series $\sum a_n$, and I cut off the first 55 quintillion terms, and replace them all with $n!$ to get a new sequence

$$ b_n = \begin{cases} a_n \text{ if } n > 55,000,000,000,000,000,000 \\ n! \text{ if } n \leq 55,000,000,000,000,000,000 \end{cases}$$

Then the sum $\sum b_n$ is still convergent. In fact, $b_n$ is convergent if and only if $a_n$ is. Of course, the sums will be different, but by precisely

$$ \sum\limits_{n=1}^{\infty} b_n - \sum\limits_{n=1}^{\infty}a_n = \sum\limits_{n=1}^{55,000,000,000,000,000,000} (n! - a_n)$$

Which is just a finite number. Of course, I've contrived the example to be huge and ridiculous. But the the point to be made about the integral test is that as long as the function behaves in the desired way beyond some large $N$ (say 55 quintillion), the argument still works for the infinite part of the sum, beyond that, and that is where all problems of convergence lie. Everything else is just addition.

  • $\begingroup$ Doesn't this mean the integral should be evaluated from $N$ to $\infty$ instead of from the original index $k$ to $\infty$ for it to be accurate? $\endgroup$
    – MrCMedlin
    Commented Jul 14, 2014 at 15:45
  • 1
    $\begingroup$ Again, as long as the integral is of a nice (read continuous) function on the interval $[0, \infty)$. The convergence of the integral is focused in the tail as we can write $$\int_0^\infty f dx = \int_0^N f dx + \int_N^\infty f dx$$ And all of the convergence/divergence behavior of the improper integral is concentrated in the second term on the right. The first term in the right is (assuming the function is continuous up to and including 0), simply some finite integral. This is admittedly a little more subtle. $\endgroup$
    – jxnh
    Commented Jul 14, 2014 at 16:18
  • $\begingroup$ OK, this makes a lot more sense now. They just gloss over this in the course without really explaining it ... you've helped a bunch! $\endgroup$
    – MrCMedlin
    Commented Jul 14, 2014 at 16:36

I do not believe your definition of the test is correct. Citing the ever-accurate (tongue-in-cheek) Wikipedia, we find the definition to be:

Consider an integer $N$ and a non-negative function $f$ defined on the unbounded interval $[N, ∞)$, on which it is monotone decreasing. Then the infinite series $$\sum_{n=N}^\infty f(n)$$ converges to a real number if and only if the improper integral $$\int_N^\infty f(x)\,dx$$ is finite. In other words, if the integral diverges, then the series diverges as well.

(Source: http://en.wikipedia.org/wiki/Integral_test_for_convergence)

It is important to note that various authors of textbooks may define a test or theorem in the way that it best fits their course outline. This is particularly true of elementary texts (e.g. first or second year calculus). Remember that mathematicians also like to generalize--if you can take a specific formulation of a theorem and generalize it (e.g. "$f$ must be decreasing everywhere" to "$f$ only has to have decreasing end-behavior"), that's perfectly acceptable.

As a further note in response to your comment on the main question: "decreasing a majority of the time" is different than "decreasing beyond $x=N$." The latter is the option you want.

  • $\begingroup$ Thank you for your response! I've reviewed the Wikipedia entry on the topic as part of my earlier research. You'll note at the beginning it says the function must be monotone decreasing on the interval being considered. This means that for the series $a_n$, $a_1 \ge a_2 \ge a_3 \ge a_4 \ge ... \ge a_n$. We can test for this by looking at the behavior of the first derivative of the function. If the first derivative is ever positive for any value $x$ in our domain, this indicates an increase from one term to the next, which would mean the function is not monotone decreasing as required. $\endgroup$
    – MrCMedlin
    Commented Jul 14, 2014 at 4:08

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