# What does it mean for a sequence to be decreasing in regards to the Integral Test for Convergence?

Example: $$\sum \limits_{n=1}^{\infty} n^k e^{-n}$$

This is the problem that I am to solve with the Integral test. I know that it converges, and I know the answer, but just don't fully get the point about it being decreasing.

How does one if a function is decreasing, the first derivative is less than zero.

$f^{\prime}\left(x\right) = \dfrac{\left(k-x\right)x^{k-1}}{e^x}$ is less than zero when $x > k$. I do understand that there is always an $x > k$. (Not really sure if what happens when $k = \infty$ leaving that issue till I learn more).

So my confusion comes to when can you say that a sequence is decreasing? What about when $x < k$: does the Integral test fail on that interval? Or just does the long-term behavior is all that matters? I mean what if just the very last number before infinity the sequence decreases, then will the Integral Test for convergence work?

Could you explain in the realm of sequences and series what does it truly mean to be decreasing? I do understand that I am just learning this, but don't be afraid to start at a high school level and build up your answers to a Field Medal level of understanding. :)

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A sequence $a_n$ is decreasing if $a_{n+1} \le a_n$ for all $n$. –  Javier Mar 5 '13 at 13:54
@JavierBadia So what about the example I give, if $k = 3$ then for the first three terms its is not decreasing? Why doesn't that invalidate the Integral Test for Convergance? –  yiyi Mar 5 '13 at 13:55
The first terms don't affect convergence. You want to know what happens for $n$ large. –  L. F. Mar 5 '13 at 13:57
@L.F. Why don't the first terms don't affect convergance? That is really my question. Thanks for putting it into words. –  yiyi Mar 5 '13 at 13:57
Because there are only finitely many first terms, so their sum will always converge. As long as your sequence is defined, it can jump, fly, or hula hoop at the beginning for all that matters. What we really want to know, is if the "infinite portion" (i.e. what happens after a certain point) will sum to a value or not. –  L. F. Mar 5 '13 at 14:01

A sequence is of the form $\{x_n\}$, and a series is of the form $\sum \limits_{n=1}^{\infty} x_n$. So for a sequence $\{x_n\}$ to be decreasing means $x_1 > x_2 > x_3 > \ldots$.

As long as there is some $k$ such that $f(n)$ is decreasing for all $n > k$, the integral test applies.

Consider the series $\sum \limits_{n=1}^{\infty} f(n) = \sum \limits_{n=1}^k f(n) + \sum \limits_{n=k + 1}^{\infty} f(n)$. The first term is finite, so converges, and the second term converges due to the integral test because it is always decreasing.

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So what about the example I give, if k=3 then for the first three terms its is not decreasing? Why doesn't that invalidate the Integral Test for Convergance? –  yiyi Mar 5 '13 at 13:56
The integral test applies as long as the function is eventually decreasing. –  ferson2020 Mar 5 '13 at 13:59
Thanks for your answer, but it has never been explained why the long term behavior is only which is important? Is it because if it converges then it stops "growing/shrinking" at that number, otherwise it just "grows/decreases" without limit? –  yiyi Mar 5 '13 at 14:03
I added to my answer; does that help make it clear? –  ferson2020 Mar 5 '13 at 14:04

Say a sequence $\{a_n\}_{n\in\mathbb{N}}$ is decreasing provided $a_{n+1}\leq a_n$ for every $n\in\mathbb{N}$. So the bigger subscripts correspond to smaller numbers. You really just need it decreasing from a rank on, so decreasing for large $n\in\mathbb{N}$.

You are looking at a series. What you want to show is that the associated sequence (that is, given the series $\sum_{n=1}^\infty a_n$ the associated sequence is $\{a_n\}_{n\in\mathbb{N}}$) is nonnegative and decreasing, then you define a function $f:[1,+\infty)\to[0,+\infty)$ so that $f(n)=a_n$ and $\int_1^\infty f<+\infty$.

Also, $k$ is an exponent so it can't be $\infty$.

Also, for series we only care what happens "near $\infty$", so to speak. So say if your sequence doesn't start decreasing until $n=100$ then just apply the integral test to the series $\sum_{n=100}^\infty a_n$, in which case you look at $\int_{100}^\infty f$.

Rationale for "near $\infty$": First of all, usually by something happening "near $\infty$ we mean there exists $N\in\mathbb{N}$ so that for all $x\geq N$ the condition happens. Convergence of a series (or a sequence for that matter, and keep in mind a series can be seen as a sequence of partial sums) only has to do with the "tail," or what happens near $\infty$. This is because for any natural number $N\in\mathbb{N}$ the first part of the series $\sum_{n=1}^Na_n$ is always finite. No matter what $N$ you choose. Because of this, if $\sum_{n=1}^\infty$ is going to diverge the problem must occur near $\infty$.

So if your associated sequence doesn't start decreasing until after $n=10,000,000,000$, no problem. Just write $\sum_{n=1}^\infty a_n=\sum_{n=1}^{10,000,000,000}a_n+\sum_{n=10,000,000,000}^\infty a_n$ and apply the integral test to the tail.

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Nice and informative answer, could you point me to some papers which explain the rational why only "near $\infity$" matters? –  yiyi Mar 5 '13 at 14:01