Integral Test for Series Convergence

The integral test is a practical tool for determining whether a series with positive or eventually positive terms converges or diverges. It applies when the general term forms a decreasing sequence that approaches zero.

How it works

Start by checking that the series has positive or eventually positive terms:

$$ \sum_{k=0}^{\infty} a_k $$

Note. A series is eventually positive if there exists an index k₀ such that all terms are positive for every k ≥ k₀. The first few terms may still be zero or negative.

Next, verify that the general term defines a decreasing sequence {a_k} that tends to zero:

$$ \lim_{k \rightarrow \infty} a_k = 0 $$

If both conditions are satisfied, the integral test can be applied.

Associate the sequence {a_k} with a continuous function f(x):

$$ \{ a_k \} \rightarrow f(x) $$

Then compute the integral of the associated function from k₀ to k, and define the auxiliary sequence {t_k}:

$$ \{ t_k \} = \int_{k_0}^k f(x) \, dx $$

Note. Here, k₀ is the starting index of the series. For example, if the series begins at k = 0, then k₀ = 0; if it begins at k = 1, then k₀ = 1.

The key idea is simple: the auxiliary sequence {t_k} and the original series share the same behavior.

• If {t_k} diverges, the series diverges.
• If {t_k} converges, the series converges.

Note. Since the series has positive terms, it cannot oscillate. It either converges or diverges.

An example

Consider the series

$$ \sum_{k=1}^{\infty} \frac{1}{\sqrt{k}} $$

This is a series with positive terms.

Note. The series satisfies the necessary condition for convergence, since $$ \lim_{k \rightarrow \infty} \frac{1}{\sqrt{k}} = 0 $$ However, this condition alone is not enough to determine whether the series converges.

The sequence {a_k} is decreasing:

$$ \left\{ \frac{1}{\sqrt{k}} \right\} \downarrow $$

It also approaches zero:

$$ \lim_{k \rightarrow \infty} \frac{1}{\sqrt{k}} = 0 $$

Both conditions required for the integral test are satisfied.

Now define the associated function:

$$ \left\{ \frac{1}{\sqrt{k}} \right\} \rightarrow f(x) = \frac{1}{\sqrt{x}} $$

Compute the integral from 1 to k to obtain the auxiliary sequence:

$$ \{ t_k \} = \int_1^k \frac{1}{\sqrt{x}} \, dx $$

Rewrite the integrand:

$$ \{ t_k \} = \int_1^k x^{-\frac{1}{2}} \, dx $$

Now integrate:

$$ \{ t_k \} = \int_1^k x^{-\frac{1}{2}} \, dx = \left[ \frac{x^{1-\frac{1}{2}}}{1-\frac{1}{2}} \right]_1^k $$

$$ \{ t_k \} = \frac{k^{1-\frac{1}{2}}}{1-\frac{1}{2}} - \frac{1^{1-\frac{1}{2}}}{1-\frac{1}{2}} $$

$$ \{ t_k \} = \frac{k^{1-\frac{1}{2}} - 1^{1-\frac{1}{2}}}{1-\frac{1}{2}} $$

$$ \{ t_k \} = \frac{k^{\frac{2-1}{2}} - 1^{\frac{2-1}{2}}}{\frac{2-1}{2}} $$

$$ \{ t_k \} = \frac{k^{\frac{1}{2}} - 1^{\frac{1}{2}}}{\frac{1}{2}} $$

$$ \{ t_k \} = \frac{\sqrt{k} - \sqrt{1}}{\frac{1}{2}} $$

$$ \{ t_k \} = 2 \cdot (\sqrt{k} - 1) $$

This is the auxiliary sequence {t_k}.

Now analyze its behavior as k → ∞.

The auxiliary sequence diverges:

$$ \lim_{k \rightarrow \infty} 2 \cdot (\sqrt{k} - 1) = +\infty $$

Therefore, by the integral test, the original series also diverges:

$$ \sum_{k=1}^{\infty} \frac{1}{\sqrt{k}} = +\infty $$

From a graphical point of view, the series shows a clear divergent trend:

Note. This is consistent with the fact that the series is a generalized harmonic series with exponent between 0 and 1, which is known to diverge.

And so on.