Integral test

Consider a series \(\sum_{"{n=1}"}^∞ a_n\). Let there be an integer \(N\) and a monotone decreasing function \(f\), such that \(f\) is defined on \([N, ∞)\) and \(f(n) = a_n\) for \(n \ge N\).

Suppose \(N=2\), and suppose \(f\) be monotone decreasing when \(x \gt 0\), then the left Riemann sum of \(\int^6_2 f(x) dx\) would be the sum of the rectangles shown below.

The area of the recantagle between 2 and 3 is \(f(2)*1 = a_2 *1 = a_2\), the area of the recantagle between 3 and 4 is \(f(3)*1 = a_3 *1 = a_3\), and so on.

\(\sum_{"{n=2}"}^5 a_n\) would be an overestimate of \(\int^6_2 f(x) dx\). This means:

\[ \sum_{n=2}^5 a_n \ge \int^6_2 f(x) dx \]

The right Riemann sum of (int^5_2 f(x) dx) would be the sum of the rectangles shown below:

\(\sum_{"{n=3}"}^5 a_n\) would be an underestimate of \(\int^5_2 f(x) dx\). This means:

\[ \begin{gather} \sum_{n=3}^5 a_n \le \int^5_2 f(x) dx \\ \sum_{n=2}^5 a_n \le a_2 + \int^5_2 f(x) dx \end{gather} \]

Putting them together:

\[ \int^6_2 f(x) dx \le \sum_{n=2}^5 a_n \le a_2 + \int^5_2 f(x) dx \]

We can generalize this:

\[ \int^{k+1}_N f(x) dx \le \sum_{n=N}^k a_n \le a_n + \int^k_N f(x) dx \]

If we take \(k \to \infty\):

\[ \int^\infty_N f(x) dx \le \sum_{n=N}^\infty a_n \le a_n + \int^\infty_N f(x) dx \]

Integral Test