<Home>Teaching><MATH 31B (Fall 2020)>Quiz 3 Solutions

Quiz 2 Solutions

Table of Contents

Question 1

If f(x)=∑n=0∞cnxn=0f\p{x} = \displaystyle\sum_{n=0}^\infty c_nx^n = 0 for all xx in an interval (−a,a)\p{-a, a}, then cn=0c_n = 0 for all nn.

Solution.

True.

The coefficients are given by cn=f(n)(0)n!c_n = \displaystyle \frac{f^{\p{n}}\p{0}}{n!}, so if f(x)=0f\p{x} = 0, then f(n)(0)=0f^{\p{n}}\p{0} = 0 for all n≥1n \geq 1. Thus, cn=0c_n = 0 for all n≥1n \geq 1.

Question 2

If an>0a_n > 0 and ∑nan\displaystyle \sum_n a_n converges, then ∑nan2\displaystyle \sum_n a_n^2 converges.

Solution.

True.

Since ∑nan\displaystyle \sum_n a_n converges, we know that lim⁡n→∞an=0\displaystyle \lim_{n\to\infty} a_n = 0. This means there exists M>0M > 0 such that if n≥Mn \geq M, then an≤1  ⟹  0≤an2≤ana_n \leq 1 \implies 0 \leq a_n^2 \leq a_n. Thus, ∑nan2\displaystyle \sum_n a_n^2 converges by direct comparison.

Question 3

Find all possible values of xx for which the series ∑n=0∞9+xn5n\displaystyle \sum_{n=0}^\infty \frac{9 + x^n}{5^n} converges.

Solution.

−5<x<5-5 < x < 5.

Since ∑n=0∞xn5n\displaystyle \sum_{n=0}^\infty \frac{x^n}{5^n} is a geometric series, it converges if and only if ∣x∣<5\abs{x} < 5, so we need to check what happens if ∣x∣≥5\abs{x} \geq 5.

If x=5x = 5, then

lim⁡n→∞9+5n5n=1≠0,\lim_{n\to\infty} \frac{9 + 5^n}{5^n} = 1 \neq 0,

so the series diverges in this case. Similarly, if x=−5x = -5, then

9+(−5)n5n=95n+(−1)n,\frac{9 + \p{-5}^n}{5^n} = \frac{9}{5^n} + \p{-1}^n,

which doesn't converge, so the series diverges in this case, too. Finally, if ∣x∣>5\abs{x} > 5, then

lim⁡n→∞9+xn5n=∞,\lim_{n\to\infty} \frac{9 + x^n}{5^n} = \infty,

so the series diverges when ∣x∣≥5\abs{x} \geq 5. This means that the interval of convergence is −5<x<5-5 < x < 5.

Question 4

Which of the following alternating series converge conditionally, but not absolutely?

  1. ∑n=2∞(−1)nnln⁡n\displaystyle\sum_{n=2}^\infty \frac{\p{-1}^n}{\sqrt{n}\ln{n}}
  2. ∑n=2∞(−1)nn(ln⁡n)2\displaystyle\sum_{n=2}^\infty \frac{\p{-1}^n}{n\p{\ln{n}}^2}
  3. ∑n=1∞cos⁡(πn)2n−3\displaystyle\sum_{n=1}^\infty \frac{\cos\p{\pi n}}{2^{n-3}}
Solution.

Only (1).

It should be easy to see that these all converge conditionally by the alternating series test. For the third one, cos⁡(πn)=(−1)n\cos\p{\pi n} = \p{-1}^n, which is why the test applies.

  1. If nn is big enough, then ln⁡n≤n\ln{n} \leq \sqrt{n}. If we multiply both sides by n\sqrt{n}, we get nln⁡n≤n\sqrt{n}\ln{n} \leq n for large nn, so

    0≤1n≤1nln⁡n.0 \leq \frac{1}{n} \leq \frac{1}{\sqrt{n}\ln{n}}.

    The smaller series diverges by pp-series, so by direct comparison, the larger series diverges, which is why (1) is conditionally convergent.

  2. In this case, the integral test applies. If we set u=ln⁡xu = \ln{x}, then du=dxx\diff{u} = \frac{\diff{x}}{x}, which gives

    ∫2∞dxx(ln⁡x)2=∫ln⁡2∞duu2,\int_2^\infty \frac{\diff{x}}{x\p{\ln{x}}^2} = \int_{\ln{2}}^\infty \frac{\diff{u}}{u^2},

    which converges by pp-integrals. By the integral test, (2) converges absolutely.

  3. If we take absolute values, then the series is the geometric series

    ∑n=1∞12n−3=18∑n=1∞(12)n,\sum_{n=1}^\infty \frac{1}{2^{n-3}} = \frac{1}{8} \sum_{n=1}^\infty \p{\frac{1}{2}}^n,

    so (3) converges absolutely.

Question 5

Find the power series representation centered at 00 of the function f(x)=ln⁡(1+x)f\p{x} = \ln\p{1 + x} for ∣x∣<1\abs{x} < 1.

Solution.

We can start with the geometric series:

11−x=∑n=0∞xn.\frac{1}{1 - x} = \sum_{n=0}^\infty x^n.

If we replace xx with −x-x, then we get

11+x=∑n=0∞(−1)nxn.\frac{1}{1 + x} = \sum_{n=0}^\infty \p{-1}^nx^n.

If we integrate both sides, we get

ln⁡(1+x)=C+∑n=0∞(−1)nxn+1n+1.\ln\p{1 + x} = C + \sum_{n=0}^\infty \p{-1}^n \frac{x^{n+1}}{n+1}.

Plugging in x=0x = 0, we get 0=ln⁡1=C0 = \ln{1} = C, so our final answer is

ln⁡(1+x)=∑n=0∞(−1)nxn+1n+1=x−x22+x33−x44+⋯.\ln\p{1 + x} = \boxed{\sum_{n=0}^\infty \p{-1}^n \frac{x^{n+1}}{n+1} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots}.