The following function
is defined for all
, and
is a positive integer. Prove or provide a counterexample to the following statement.
The convergence of the improper integral
![Rendered by QuickLaTeX.com \[ \int_1^{\infty} f(x) \, dx \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-8ba2febfe70755669cc3aa864a8bcd2b_l3.png)
implies
![Rendered by QuickLaTeX.com \[ \lim_{x \to \infty} f(x) = 0. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-4c8bfd5522f3d658deaed83b70cafcc6_l3.png)
Counterexample. The idea of the construction is a function which has rapidly diminishing area, but has a height that is not going to 0. (So, for an idea consider triangles on the real line all with height 1, but for which the base is becoming small rapidly.) To make this concrete, define
![Rendered by QuickLaTeX.com \[ f(x) = 1 - 2n^2 \left| x - \left( n + \frac{1}{2n^2} \right) \qquad \text{for } x \in [n,n+1), \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-384b9ed6b01d4ae5f6bc16bc2f031b45_l3.png)
for each positive integer
. Then for the improper integral we have
![Rendered by QuickLaTeX.com \[ \int_1^{\infty} f(x) \, dx \leq \sum_{n=1}^{\infty} \frac{1}{2n^2} \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-7554d4d0d4292b3044e5d8209d517301_l3.png)
which we know converges. On the other hand
![Rendered by QuickLaTeX.com \[ f \left( n + \frac{1}{2n^2} \right) = 1 \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-000639bb40630d01a5a6cfd03829a0ad_l3.png)
for all positive integers
. Hence,
![Rendered by QuickLaTeX.com \[ \lim_{x \to \infty} f(x) \neq 0 \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-417ede15cd65eb4e137111f71a6e3f14_l3.png)
(since it does not exist). Hence, the statement is false.
(Note: For more on this see this question on Math.SE.)