Solutions to Problem Set 16

Math 211-03

10-16-2017

[Sequences]

1. Determine the limits of the geometric sequences in the table below, Your answers should be 0, 1, $\infty$ , or "undefined".

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & $a_n$ & & Limit & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & $0.79^n$ & & \vphantom{$\dfrac{x}{x}$}\hphantom{xxxxx} & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & $1^n$ & & \vphantom{$\dfrac{x}{x}$}\hphantom{xxxxx} & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & $\left(\dfrac{17}{13}\right)^n$ & & \vphantom{$\dfrac{x}{x}$}\hphantom{xxxxx} & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & $(-2)^n$ & & \vphantom{$\dfrac{x}{x}$}\hphantom{xxxxx} & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} }} $$

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & $a_n$ & & Limit & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & $0.79^n$ & & 0 & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & $1^n$ & & 1 & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & $\left(\dfrac{17}{13}\right)^n$ & & $\infty$ & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & $(-2)^n$ & & Undefined & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} }}\quad\halmos $$


2. Determine whether the sequence $a_n = \dfrac{3}{n^4 + 1}$ defined for $n \ge 1$ is (eventually) increasing, decreasing, or neither.

Let $f(x) = \dfrac{3}{x^4 + 1}$ . Then

$$f'(x) = \dfrac{-12 x^3}{(x^4 + 1)^2}.$$

For $x \ge 1$ , I have $f'(x) =
   \dfrac{-12 x^3}{(x^4 + 1)^2} < 0$ . Therefore, the sequence is decreasing.


3. Determine whether the sequence $a_n = \dfrac{e^n}{n^2}$ defined for $n > 2$ is (eventually) increasing, decreasing, or neither.

Let $f(x) = \dfrac{e^x}{x^2}$ . Then

$$f'(x) = \dfrac{x^2 e^x - 2 x e^x}{x^4} = \dfrac{(x - 2) e^x}{x^3}.$$

For $x > 2$ , I have $x - 2
   > 0$ , and $e^x$ and $x^3$ are positive. Therefore, $f'(x) > 0$ for $x > 2$ , and the sequence is (eventually) increasing.


4. Determine whether the sequence $a_n = \cos n$ defined for $n > 0$ is (eventually) increasing, decreasing, or neither.

Since $\cos n$ oscillates as $n \to \infty$ , the sequence is neither eventually increasing nor eventually decreasing.


5. Prove that the sequence $a_n
   = \dfrac{7 n + 2}{5 n - 3}$ is bounded.

If a sequence has a (finite) limit, then it must be bounded.

$$\lim_{n \to \infty} \dfrac{7 n + 2}{5 n - 3} = \dfrac{7}{5}.$$

Since the sequence has a finite limit, it is bounded.


6. A sequence is defined recursively by

$$a_1 = 9, \quad a_{n + 1} = \sqrt{2 a_n + 15} \quad\hbox{for}\quad n \ge 1.$$

Find $\displaystyle \lim_{n \to
   \infty} a_n$ , assuming that the limit exists.

Let

$$L = \lim_{n \to \infty} a_n.$$

Note that

$$L = \lim_{n \to \infty} a_{n + 1}.$$

Reason: Both of these limits give the value that the a's approach, which is L:

$$a_1, \quad a_2, \quad a_3, \quad a_4, \ldots \to L.$$

Take the limit on both sides of the the recursion equation:

$$\eqalign{ \lim_{n \to \infty} a_{n + 1} & = \lim_{n \to \infty} \sqrt{2 a_n + 15} \cr \lim_{n \to \infty} a_{n + 1} & = \sqrt{2 \lim_{n \to \infty} a_n + 15} \cr L & = \sqrt{2 L + 15} \cr L^2 & = 2 L + 15 \cr L^2 - 2 L - 15 & = 0 \cr (L - 5)(L + 3) & = 0 \cr L = 5 & \quad\hbox{or}\quad L = -3 \cr}$$

The definition of the a's shows that all of them are positive, so their limit can't be negative. Therefore,

$$\lim_{n \to \infty} a_n = 5.\quad\halmos$$


[Series]

7. Determine whether the series $\displaystyle \sum_{n=0}^\infty \left(\dfrac{4}{5}\right)^n$ converges or diverges. If the series converges, find the exact value of its sum.

The series is geometric, with ratio $\dfrac{4}{5}$ . Hence, it converges, and its sum is

$$\dfrac{1}{1 - \dfrac{4}{5}} = 5.\quad\halmos$$


8. Determine whether the series $\displaystyle \sum_{n=1}^\infty 4 \cdot (-0.9)^n$ converges or diverges. If the series converges, find the exact value of its sum.

The series is geometric, with ratio -0.9. Hence, it converges, and its sum is

$$\dfrac{4 \cdot (-0.9)}{1 - (-0.9)} = -\dfrac{36}{19}.\quad\halmos$$


9. Determine whether the series $\displaystyle \sum_{n=0}^\infty \dfrac{11^n}{7^n}$ converges or diverges. If the series converges, find the exact value of its sum.

The series is geometric, with ratio $\dfrac{11}{7}$ . Hence, it diverges.


10. Determine whether the following series converges or diverges. If the series converges, find the exact value of its sum.

$$\dfrac{6}{5^2} + \dfrac{6}{5^3} + \cdots + \dfrac{6}{5^n} + \cdots .$$

The series is geometric, with ratio $\dfrac{1}{5}$ . Hence, it converges, and its sum is

$$\dfrac{\dfrac{6}{5^2}}{1 - \dfrac{1}{5}} = \dfrac{3}{10}.\quad\halmos$$


11. Determine whether the series $\displaystyle \sum_{n=1}^\infty \left(\dfrac{3^n}{7^n} +
   \dfrac{1}{3^n}\right)$ converges or diverges. If the series converges, find the exact value of its sum.

The series is the sum of two geometric series.

The series $\displaystyle
   \sum_{n=1}^\infty \dfrac{3^n}{7^n}$ has ratio $\dfrac{3}{7}$ , so it converges. Its sum is

$$\dfrac{\dfrac{3}{7}}{1 - \dfrac{3}{7}} = \dfrac{3}{4}.$$

The series $\displaystyle
   \sum_{n=1}^\infty \dfrac{1}{3^n}$ has ratio $\dfrac{1}{3}$ , so it converges. Its sum is

$$\dfrac{\dfrac{1}{3}}{1 - \dfrac{1}{3}} = \dfrac{1}{2}.$$

Hence, the original series converges, and its sum is

$$\sum_{n=1}^\infty \left(\dfrac{3^n}{7^n} + \dfrac{1}{3^n}\right) = \dfrac{3}{4} + \dfrac{1}{2} = \dfrac{5}{4}.\quad\halmos$$


12. Write the following infinite repeating decimal as a geometric series, and find the exact value of its sum:

$$0.29292929 \ldots.$$

$$0.29292929 \ldots = \dfrac{29}{100} + \dfrac{29}{10000} + \dfrac{29}{1000000} + \cdots.$$

The series is geometric with ratio $\dfrac{1}{100}$ . Its sum is

$$\dfrac{\dfrac{29}{100}}{1 - \dfrac{1}{100}} = \dfrac{29}{99}.\quad\halmos$$


13. Determine whether the series $\displaystyle \sum_{n=1}^\infty \dfrac{7}{n}$ converges or diverges. If the series converges, find the exact value of its sum.

Note that

$$\sum_{n=1}^\infty \dfrac{7}{n} = 7 \sum_{n=1}^\infty \dfrac{1}{n}.$$

The series is 7 times the harmonic series, which diverges. So the original series diverges.


14. Use telescoping to find the exact value of the sum of the series $\displaystyle \sum_{n =
   1}^\infty \dfrac{2}{n (n + 2)}$ .

Hint: By partial fractions,

$$\dfrac{2}{n (n + 2)} = \dfrac{1}{n} - \dfrac{1}{n + 2}.$$

I have

$$\sum_{n = 1}^\infty \dfrac{2}{n (n + 2)} = \sum_{n = 1}^\infty \left(\dfrac{1}{n} - \dfrac{1}{n + 2}\right).$$

To see what is happening, write out the first few terms of the last sum:

$$\sum_{n = 1}^\infty \left(\dfrac{1}{n} - \dfrac{1}{n + 2}\right) =$$

$$\left(\dfrac{1}{1} - \dfrac{1}{3}\right) + \left(\dfrac{1}{2} - \dfrac{1}{4}\right) + \left(\dfrac{1}{3} - \dfrac{1}{5}\right) + \left(\dfrac{1}{4} - \dfrac{1}{6}\right) + \left(\dfrac{1}{5} - \dfrac{1}{7}\right) + \left(\dfrac{1}{6} - \dfrac{1}{8}\right) + \cdots .$$

You can see that the terms $\dfrac{1}{3}$ , $\dfrac{1}{4}$ , and so on cancel in pairs. The only terms which don't cancel are $\dfrac{1}{1}$ and $\dfrac{1}{2}$ . So

$$\sum_{n = 1}^\infty \left(\dfrac{1}{n} - \dfrac{1}{n + 2}\right) = \dfrac{1}{1} + \dfrac{1}{2} = \dfrac{3}{2}.\quad\halmos$$


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