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10.4 The Divergence and Integral Tests

How do we know if an infinite series \(\sum_{k=1}^{\infty} a_k\) converges?

We will see several "tests" to test if it converges.

Last time: geometric series

\[ \sum_{k=0}^{\infty} ar^k = a + ar + ar^2 + ar^3 + \dots \]

converges if \(|r| < 1\)

What about \(1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \dots = \sum_{k=1}^{\infty} \frac{1}{k^2}\)?

Let's look at the convergence question from the opposite point of view

→ if a series converges, what must happen?

\[ \sum_{k=1}^{\infty} a_k \]

\(S_1 = a_1\) first partial sum

\(S_2 = a_1 + a_2\)

\(S_3 = a_1 + a_2 + a_3\)

\(\vdots\)

\(S_{k-1} = a_1 + a_2 + a_3 + \dots + a_{k-1}\)

\(S_k = a_1 + a_2 + a_3 + \dots + a_{k-1} + a_k\)

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Convergence: \(\lim_{k \to \infty} S_k = L\) (a finite number)

\(\lim_{k \to \infty} S_{k-1} = L\)

From last page, note:

\(S_2 - S_1 = a_2\)

\(S_3 - S_2 = a_3\)

\(S_k - S_{k-1} = a_k\)

Then

\[ \lim_{k \to \infty} (S_k - S_{k-1}) = \lim_{k \to \infty} a_k \]

\(\lim_{k \to \infty} S_k\)

\(\underbrace{\quad\quad\quad}_{L}\)

\(-\)

\(\lim_{k \to \infty} S_{k-1}\)

\(\underbrace{\quad\quad\quad}_{L}\)

\(=\)

\(\lim_{k \to \infty} a_k = 0\)

This is the Divergence Test

If \(\sum_{k=1}^{\infty} a_k\) converges, then \(\lim_{k \to \infty} a_k = 0\)

This is a one way if. The converse is not necessarily true.

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for example, \(\sum_{k=0}^{\infty} (\frac{1}{3})^k\) is a geometric series with \(|r| < 1\) (\(r = \frac{1}{3}\)) so this converges

\[\text{notice } \lim_{k \to \infty} a_k = \lim_{k \to \infty} (\frac{1}{3})^k = \lim_{k \to \infty} \frac{1}{3^k} = 0\]

but \(\sum_{k=0}^{\infty} \cos(k\pi) = 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + \dots\)

\[\text{clearly diverges, note } \lim_{k \to \infty} a_k = \lim_{k \to \infty} \cos(k\pi)\]

DNE

(not zero)

but, just because \(\lim_{k \to \infty} a_k = 0\), it does NOT necessarily mean

\[\sum_{k=1}^{\infty} a_k \text{ converges.}\]

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\(\sum_{k=1}^{\infty} \frac{1}{k} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \dots\)"Harmonic Series"

clearly, \(\lim_{k \to \infty} a_k = \lim_{k \to \infty} \frac{1}{k} = 0\)

but does it converge?

Let's look at some partial sums

\(s_1 = 1\)

\(s_2 = 1 + \frac{1}{2} = 1.5\)

\(s_3 = 1 + \frac{1}{2} + \frac{1}{3} = 1.833\)

\(s_4 = 2.0833\)

\(\vdots\)

\(s_{20} = 3.5977\)

\(\vdots\)

\(s_{50} = 4.4992\)

\(\vdots\)

\(s_{100} = 5.1874\)

\(\vdots\)

\(s_{500} = 6.7928\)

there is no sign of \(s_k\) settling down

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Harmonic Series Partial Sums: n = 300

\[ \sum_{k=1}^{\infty} \frac{1}{k} \]

The following graph illustrates the partial sums \( S_n \) of the harmonic series for \( n \) up to 300. Note that the curve is not flattening out, suggesting that the series does not converge to a finite value.

Figure: Graph of partial sums S_n for the harmonic series from n=0 to 300, showing a logarithmic growth curve.

Key points labeled on the graph include:

\( S_1 \) at \( n=1 \)
\( S_2 \) at \( n=2 \)
\( S_3 \) at \( n=3 \)
\( S_{300} \) at the end of the plotted range, reaching a value above 6.

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Harmonic Series Partial Sums: n = 1000

\[ \sum_{k=1}^{\infty} \frac{1}{k} \]

This graph extends the visualization of the partial sums \( S_n \) of the harmonic series up to \( n = 1000 \). Even at this larger scale, the growth continues without reaching a horizontal asymptote.

Figure: Graph of partial sums S_n for the harmonic series from n=0 to 1000, showing continued slow growth.

Observations from the extended scale:

Initial partial sums \( S_1 \) and \( S_2 \) are marked near the origin.
The partial sum \( S_{1000} \) is reached at the far right, with a value between 7 and 8.

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Convergence and the Integral Test

The Harmonic Series \[ \sum_{k=1}^{\infty} \frac{1}{k} \] has \( \lim_{k \to \infty} a_k = 0 \) but does NOT converge.

So, again, just because \( \lim_{k \to \infty} a_k = 0 \) it does NOT mean \( \sum_{k=1}^{\infty} a_k \) converges.

BUT, if \( \lim_{k \to \infty} a_k \neq 0 \), then \( \sum_{k=1}^{\infty} a_k \) diverges.

So, if we know \( \lim_{k \to \infty} a_k = 0 \), how do we make sure that the series \( \sum_{k=1}^{\infty} a_k \) converges?

Today, we will see the Integral Test

\[ \sum_{k=1}^{\infty} \frac{1}{k^2} = 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \dots \]

Treat as points on graph of \( f(x) = \frac{1}{x^2} \) at \( x = 1, 2, 3, 4, \dots \)

\( \lim_{k \to \infty} \frac{1}{k^2} = 0 \), passes the Divergence Test but not guaranteed to converge.

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Graph Analysis

Look at \( 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \dots \) as a Riemann sum approximating the integral:

\[ \int_{1}^{\infty} \frac{1}{x^2} dx \]

Notice the boxes are below the curve, so the approx. is an underestimate.

Figure: Coordinate graph of f(x)=1/x^2 with red shaded rectangles showing an under-approximation Riemann sum.

Graph of \( f(x) = \frac{1}{x^2} \)

\[ \sum_{k=1}^{\infty} \frac{1}{k^2} \leq 1 + \int_{1}^{\infty} \frac{1}{x^2} dx \]

Note: 1 represents the first box.

We will revisit this.

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The Integral Test and Riemann Sums

Now we look at \(\sum_{k=1}^{\infty} \frac{1}{k^2} = 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \dots\) as a Right Riemann Sum.

The following graph illustrates the series as a sum of rectangular areas compared to the curve \(f(x) = \frac{1}{x^2}\). The rectangles represent the terms of the series, where the first rectangle has an area of \(a_1 = 1\), the second \(a_2 = \frac{1}{4}\), and so on.

Figure: Coordinate graph of f(x) = 1/x^2 with right-endpoint rectangles a1, a2, a3 showing an overestimate.

this estimate is an over estimate

\[\sum_{k=1}^{\infty} \frac{1}{k^2} \geq \int_{1}^{\infty} \frac{1}{x^2} dx\]

Combining the Bounds

Combine the two black boxes:

\[\int_{1}^{\infty} \frac{1}{x^2} dx \leq \sum_{k=1}^{\infty} \frac{1}{k^2} \leq 1 + \int_{1}^{\infty} \frac{1}{x^2} dx\]

So, if \(\int_{1}^{\infty} \frac{1}{x^2} dx\) converges, then \(\sum_{k=1}^{\infty} \frac{1}{k^2}\) also converges.

The Integral Test

this is the Integral Test.

\[\sum_{k=1}^{\infty} a_k = \sum_{k=1}^{\infty} a(k)\]

\(\hookrightarrow f(x)\)

\(\int_{1}^{\infty} f(x) dx\) if conv. then series conv.

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Convergence Analysis

So, does \(\sum_{k=1}^{\infty} \frac{1}{k^2}\) converge?

does \(\int_{1}^{\infty} \frac{1}{x^2} dx\) converge? yes, in fact \(\int_{1}^{\infty} \frac{1}{x^p} dx\) converges if \(p > 1\)

So, that means \(\sum_{k=1}^{\infty} \frac{1}{k^p}\) converges if \(p > 1\)

\(\sum_{k=1}^{\infty} \frac{1}{k^p}\)

p-series

converges if \(p > 1\)

"p-series test"

Examples

for example, \(\sum_{k=1}^{\infty} \frac{1}{k^7}\) converges because \(p = 7 > 1\)

and \(\sum_{k=1}^{\infty} \frac{1}{k}\) diverges because \(p = 1\)

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Convergence Tests for Series

The Divergence Test

The divergence test (\[ \lim_{k \to \infty} a_k = 0? \]) can save us time.

If \[ \lim_{k \to \infty} a_k \neq 0 \], no need to use Integral (or any other) test.
If \[ \lim_{k \to \infty} a_k = 0 \], then we investigate more.

The Integral Test

The Integral Test can establish convergence, but does not give us the sum (where the series converges to).

The p-series Test

A consequence of the Integral Test is the p-series Test:

\[ \sum_{k=1}^{\infty} \frac{1}{k^p} \text{ converges if } p > 1 \]

Starting Index and Convergence

The starting \( k \) does NOT affect convergence.

If \[ \sum_{k=1}^{\infty} a_k \text{ converges, then } \sum_{k=1337}^{\infty} a_k \text{ also converges} \]
If \[ \sum_{k=1}^{\infty} a_k \text{ diverges, then } \sum_{k=1337}^{\infty} a_k \text{ still diverges.} \]