Solutions to Problem Set 23

Math 211-03

11-2-2017

[Alternating series]

1. Determine whether the series $\displaystyle \sum_{n=1}^\infty (-1)^{n + 1} \dfrac{9}{n}$ converges or diverges.

The terms alternate, and

$$\lim_{n \to \infty} \dfrac{9}{n} = 0.$$

If $f(x) = \dfrac{9}{x}$ , then

$$f'(x) = -\dfrac{9}{x^2} < 0.$$

Hence, the terms decrease in absolute value.

Therefore, the series converges by the Alternating Series Test.


2. Determine whether the series $\displaystyle \sum_{n=2}^\infty (-1)^n \dfrac{1}{\sqrt{2 n +
   1}}$ converges or diverges.

The terms alternate, and

$$\lim_{n \to \infty} \dfrac{1}{\sqrt{2 n + 1}} = 0.$$

If $f(x) = \dfrac{1}{\sqrt{2 x +
   1}} = (2 x + 1)^{-1/2}$ , then

$$f'(x) = -\dfrac{1}{(2 x + 1)^{3/2}} < 0.$$

Hence, the terms decrease in absolute value.

Therefore, the series converges by the Alternating Series Test.


3. Determine whether the series $\displaystyle \sum_{n=1}^\infty (-1)^{n + 1} \dfrac{4 n + 3}{2
   n + 1}$ converges or diverges.

The terms alternate, but

$$\lim_{n \to \infty} \dfrac{4 n + 3}{2 n + 1} = 2 \ne 0.$$

The series diverges by the Zero Limit Test.


4. Determine whether the series $\displaystyle \sum_{n=2}^\infty (-1)^n \dfrac{n}{2 n^2 + 1}$ converges or diverges.

The terms alternate, and

$$\lim_{n \to \infty} \dfrac{n}{2 n^2 + 1} = 0.$$

If $f(x) = \dfrac{x}{2 x^2 +
   1}$ , then

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

Now $1 - 2 x^2 < 0$ for $x
   > \dfrac{1}{\sqrt{2}}$ , so $f'(x) < 0$ for $x \ge 2$ . Hence, the terms decrease in absolute value.

The series converges by the Alternating Series Test.


5. Determine whether the series $\displaystyle \sum_{n=1}^\infty (-1)^{n + 1} \dfrac{n}{3^n}$ converges or diverges.

The terms alternate, and by L'H\^opital's rule

$$\lim_{n \to \infty} \dfrac{n}{3^n} = \lim_{n \to \infty} \dfrac{1}{3^n \ln 3} = 0.$$

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

$$f'(x) = \dfrac{(3^x)(1) - (x)(3^x \ln 3)}{(3^x)^2} = \dfrac{1 - x \ln 3}{3^x} < 0.$$

Hence, the terms decrease in absolute value.

Therefore, the series converges by the Alternating Series Test.


6. Determine whether the series $\displaystyle \sum_{n=2}^\infty (-1)^n \dfrac{2^n}{n^2}$ converges or diverges.

The terms alternate. Using L'H\^opital's Rule, I have

$$\lim_{n \to \infty} \dfrac{2^n}{n^2} = \lim_{n \to \infty} \dfrac{2^n \ln 2}{2n} = \lim_{n \to \infty} \dfrac{2^n (\ln 2)^2}{2} = +\infty.$$

Hence,

$$\lim_{n \to \infty} (-1)^n \dfrac{2^n}{n^2} \quad\hbox{is undefined}.$$

Therefore, the series diverges by the Zero Limit Test.


7. Consider the convergent alternating series $\displaystyle \sum_{k=0}^\infty
   (-1)^k \dfrac{1}{(k + 3)^{1/5}}$ .

Find the smallest value of n for which the $n^{\rm th}$ partial sum $\displaystyle s_n = \sum_{k=0}^n (-1)^k \dfrac{1}{(k +
   3)^{1/5}}$ approximates the actual sum to within 0.1.

The partial sum $s_n$ differs from the actual sum s by no more than the absolute value of the next term $|a_{n + 1}|$ :

$$|s - s_n| < |a_{n + 1}|.$$

I can ensure that $|s - s_n| <
   0.1$ if $a_{n + 1} < 0.1$ . Then

$$|s - s_n| < |a_{n + 1}| < 0.1.$$

So

$$\eqalign{ \dfrac{1}{[(n + 1) + 3]^{1/5}} & < 0.1 \cr \noalign{\vskip2pt} \dfrac{1}{(n + 4)^{1/5}} & < 0.1 \cr \noalign{\vskip2pt} 1 & < 0.1 (n + 4)^{1/5} \cr 10 & < (n + 4)^{1/5} \cr 100000 & < n + 4 \cr 99996 & < n \cr}$$

Thus, $n = 99997$ .


8. Consider the convergent alternating series $\displaystyle \sum_{k=0}^\infty
   (-1)^k \dfrac{3^k}{k!}$ .

Find the smallest value of n for which the $n^{\rm th}$ partial sum $\displaystyle s_n = \sum_{k=0}^n (-1)^k \dfrac{3^k}{k!}$ approximates the actual sum to within 0.001.

The partial sum $s_n$ differs from the actual sum s by no more than the absolute value of the next term $|a_{n + 1}|$ :

$$|s - s_n| < |a_{n + 1}|.$$

I can ensure that $|s - s_n| <
   0.001$ if $a_{n + 1} < 0.001$ . Then

$$|s - s_n| < |a_{n + 1}| < 0.001.$$

That is,

$$\dfrac{3^{n + 1}}{(n + 1)!} < 0.001.$$

However, I can't solve this inequality algebraically. Therefore, I'll do this by trial and error by making a table. (I'm only showing some of the values.)

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & n & & $a_{n + 1} = \dfrac{3^{n + 1}}{(n + 1)!}$ & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & 7 & & $0.16272 \ldots$ & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & 8 & & $0.05424 \ldots$ & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & 9 & & $0.01627 \ldots$ & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & 10 & & $0.00443 \ldots$ & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & 11 & & $0.00110 \ldots$ & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & 12 & & $0.00025 \ldots$ & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} }} $$

I see that $|a_{n + 1}| <
   0.001$ for the first time when $n = 12$ . So I need to use $\displaystyle \sum_{n=0}^{12} (-1)^n \dfrac{3^n}{n!}$ to estimate the sum to within 0.001.


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