Wemazi

Topic 10.3 · BC

Find the limit of the terms

Builds on finding a sequence limit. 6 minutes.

Where do the terms head?

A series adds up the terms \(a_1, a_2, a_3, \dots\). Each dot shows one term at position \(n\), and its height is \(a_n=\frac{2n+1}{n+3}\). The orange line marks the height at the position you choose.

Drag \(n\) to the right. The height is \(\frac34\) at \(n=1\), about 1.62 at \(n=10\) and about 1.85 at \(n=30\), so the dots rise more slowly at each step.

The button draws a dashed line at height 2. The dots stay below it, and the far dots come closer to it.

The limit of the terms is the number that the heights approach when \(n\) is large. Here the limit is 2, not 0.

Three ways the dots can end

The terms now follow any rule. Press each button, then drag \(n\) far to the right and read the far dots, which are the dots at large \(n\).

If the dots rise without end, no number is close to their heights, so the limit is infinite.

If the dots approach one height, every far dot is close to it, and that height is the limit.

If the dots alternate between two heights, no single number is close to every far dot, so the limit does not exist.

Only the far dots decide, so the first dots can lie anywhere without changing the limit.

Notation

  1. The limit of the terms is the number that \(a_n\) approaches as \(n\) grows.

    \[\lim_{n\to\infty}\cy{a_n}=\cy{L}\]

    The numbers \(a_n\) are the terms you add, not the partial sums, which are their running totals.

  2. If \(a_n\) is a fraction of polynomials in \(n\), divide the top and the bottom by the highest power of \(n\) in the bottom. Each leftover fraction approaches 0, because its top stays fixed while its bottom grows.

    \[\frac{2n+1}{n+3}=\frac{2+\cq{1/n}}{1+\cq{3/n}}\to2\]

  3. If the signs alternate, first find the limit of the size \(|a_n|\), which is the term without its sign. If the size approaches 0, then \(a_n\) approaches 0. If the size approaches a number that is not 0, the limit of \(a_n\) does not exist.

    \[\lim_{n\to\infty}|a_n|=0\ \Rightarrow\ \lim_{n\to\infty}a_n=0\]

When it fails

Multiply each term by \((-1)^n\). Each term keeps its size, but the dots with odd \(n\) move below the axis. They approach height \(-2\), while the dots with even \(n\) approach height 2.

No single number is close to all the far dots, so the limit of \(a_n\) does not exist. The limit of the size \(|a_n|\) is still 2.

Example 1: A fraction of polynomials

Find \(\displaystyle\lim_{n\to\infty}a_n\) for \(a_n=\dfrac{5n^2+1}{2n^2+n}\).

  1. The highest power of \(n\) in the bottom is \(n^2\). Divide the top and the bottom by \(n^2\), which does not change the value of the fraction.

    \[a_n=\frac{5+\cq{1/n^2}}{2+\cq{1/n}}\]

  2. When \(n\) is large, \(1/n^2\) and \(1/n\) are close to 0, because their bottoms grow.

    \[\frac{5+\cq{1/n^2}}{2+\cq{1/n}}\to\frac{5+0}{2+0}\]

  3. The limit is \(\frac52\), and the far dots lie close to the dashed line at height 2.5.

    \[\lim_{n\to\infty}a_n=\cy{\tfrac52}\]

Example 2: A sign that alternates

Find \(\displaystyle\lim_{n\to\infty}a_n\) for \(a_n=(-1)^n\,\dfrac{3n-2}{n+4}\).

  1. The sign alternates, so find the limit of the size first. Divide the top and the bottom by \(n\).

    \[|a_n|=\frac{3-\cq{2/n}}{1+\cq{4/n}}\to3\]

  2. A tempting wrong move: The size approaches 3, so the limit of \(a_n\) is 3.

    \[\lim_{n\to\infty}a_n=3\]

    The dot at \(n=40\) has height about 2.68, but the dot at \(n=39\) has height about \(-2.67\), far below the red line at 3. The size loses the sign of each term.

  3. The dots with even \(n\) approach 3, and the dots with odd \(n\) approach \(-3\). No single number is close to all the far dots, so the limit does not exist.

Practice

Find \(\displaystyle\lim_{n\to\infty}a_n\) for \(a_n=\dfrac{4n}{\sqrt{n^2+9}}\).

Divide the top by \(n\). Divide the bottom by \(n\) as well, which moves \(n\) inside the root as \(n^2\).

Solution

  1. For \(n>0\), \(n=\sqrt{n^2}\). Divide the top and the bottom by \(n\).

    \[a_n=\frac{4}{\sqrt{1+9/n^2}}\]

  2. The fraction \(9/n^2\) approaches 0, so the bottom approaches \(\sqrt1=1\).

    \[\sqrt{1+9/n^2}\to1\]

  3. The limit is 4.

    \[\lim_{n\to\infty}a_n=4\]

Questions

Questions 1 and 2 use the figure. The upper graph shows the dots of the sequence \(a_n=4-\frac{8}{n+1}\), and the lower graph shows the dots of \(b_n=(-1)^n\,\frac{n+1}{n}\), for \(n=1\) to 30.

  1. What number do the heights of the dots in the upper graph approach as \(n\) grows?

    • 4
    • 3.74
    • 0
    • There is no limit.
    Answer

    4

  2. What is true of the limit of \(b_n\) in the lower graph?

    • It does not exist.
    • It is 0.
    • It is 1.
    Answer

    It does not exist.

  3. Find the limit of \(a_n=\dfrac{(n+4)^2}{2n^2+n}\).

    • \(\tfrac12\)
    • 1
    • 0
    • It does not exist.
    Answer

    \(\tfrac12\)

Summary

The limit of the terms is the one height that all the far dots approach, and if there is none, the limit does not exist.

For a fraction of polynomials, you divide by the highest power of \(n\) in the bottom, and for alternating signs, you find the size first.

You find this limit first for any series, because it tells you which conclusion about the series can come next.

This is one of four skills in topic 10.3 of Which & Why: AP Calculus.