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The Integral Test for Positive Series Examples 1
Let $f(n) = a_n$ be a continuous ultimately non-increasing positive function for the interval $[N, \infty)$ where $N \in \mathbb{N}$. Recall from The Integral Test for Positive Series page that if $\int_{N}^{\infty} f(x) \: dx$ converges then $\sum_{n=N}^{\infty} a_n$ converges. Similarly, if $\int_{N}^{\infty} f(x) \: dx$ diverges to infinity then $\sum_{n=N}^{\infty} a_n$ diverges to infinity.
We will now look at some examples of applying the integral test.
Example 1
Using the integral test, determine whether the series $\sum_{n=1}^{\infty} \frac{1}{1 + 10n}$ is convergent or divergent.
The answer to this question should be fairly obvious using the divergence theorem (since $\lim_{n \to \infty} a_n = 1$), but let's show it with the integral test. Let $f(x) = \frac{1}{1 + 10x}$. This function is positive for $x ≥ 1$, non-increasing, and continuous. Let $u = 1 + 10x$ and so $du = 10 \: dx$ which implies $\frac{du}{10} = dx$, and so:
(1)Therefore it follows that:
(2)Therefore the series $\sum_{n=1}^{\infty} \frac{1}{1 + 10n}$ is divergent.
Example 2
Using the integral test, determine whether the series $\sum_{n=1}^{\infty} \frac{n}{1 + n^2}$ is convergent or divergent.
First let $f(x) = \frac{x}{1 + x^2 }$. This function is positive for $x ≥ 1$, non-increasing, and continuous. We now evaluate the following integral by letting $u = 1 + x^2$ so that $du = 2x \: dx$ which implies that $\frac{du}{2} = x \: dx$, and so:
(3)Therefore it follows that:
(4)Example 3
Using the integral test, determine whether the series $\sum_{n=1}^{\infty} ne^{-n}$ is convergent or divergent.
First let $f(x) = xe^{-x}$. This function is positive for $x ≥ 1$, non-increasing, and continuous. We now evaluate the following integral with integration by parts. Let $u = x$ and let $dv = e^{-x} \: dx$ and so $du = dx$ and $v = -e^{-x}$
(5)And therefore:
(6)We now want to evaluate $\lim_{x \to \infty} \frac{x + 1}{e^x}$ using L'Hospital's rule, and so:
(7)Therefore $- \lim_{b \to \infty} \left ( \frac{b + 1}{e^b} - \frac{2}{e} \right ) = \frac{2}{e}$ and therefore $\sum_{n=1}^{\infty} ne^{-n}$ is convergent.