Power Series and Functions

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

1. For each of the following series, find the interval and radius of convergence.

   1. \(\sum _{n=0}^{\infty }\frac{{x}^{n}}{n\text{!}}\)
   2. \(\sum _{n=0}^{\infty }n\text{!}{x}^{n}\)
   3. \(\sum _{n=0}^{\infty }\frac{{(x-2)}^{n}}{(n+1){3}^{n}}\)
   Kusonyeza yankho

   1. To check for convergence, apply the ratio test. We have
      \[\begin{array}{ll}\rho & =\underset{n\to \infty }{\text{lim}}|\frac{\frac{{x}^{n+1}}{(n+1)\text{!}}}{\frac{{x}^{n}}{n\text{!}}}| \\ & =\underset{n\to \infty }{\text{lim}}|\frac{{x}^{n+1}}{(n+1)\text{!}}\cdot \frac{n\text{!}}{{x}^{n}}| \\ & =\underset{n\to \infty }{\text{lim}}|\frac{{x}^{n+1}}{(n+1)\cdot n\text{!}}\cdot \frac{n\text{!}}{{x}^{n}}| \\ & =\underset{n\to \infty }{\text{lim}}|\frac{x}{n+1}| \\ & =|x|\underset{n\to \infty }{\text{lim}}\frac{1}{n+1} \\ & =0<1\end{array}\]
      for all values of x. Therefore, the series converges for all real numbers x. The interval of convergence is \((\text{-}\infty ,\infty )\) and the radius of convergence is \(R=\infty .\)
   2. Apply the ratio test. For \(x\ne 0,\) we see that
      \[\begin{array}{ll}\rho & =\underset{n\to \infty }{\text{lim}}|\frac{(n+1)\text{!}{x}^{n+1}}{n\text{!}{x}^{n}}| \\ & =\underset{n\to \infty }{\text{lim}}|(n+1)x| \\ & =|x|\underset{n\to \infty }{\text{lim}}(n+1) \\ & =\infty .\end{array}\]
      Therefore, the series diverges for all \(x\ne 0.\) Since the series is centered at \(x=0,\) it must converge there, so the series converges only for \(x=0.\) The interval of convergence is the single value \(x=0\) and the radius of convergence is \(R=0.\)
   3. In order to apply the ratio test, consider
      \[\begin{array}{ll}\rho & =\underset{n\to \infty }{\text{lim}}|\frac{\frac{{(x-2)}^{n+1}}{(n+2){3}^{n+1}}}{\frac{{(x-2)}^{n}}{(n+1){3}^{n}}}| \\ & =\underset{n\to \infty }{\text{lim}}|\frac{{(x-2)}^{n+1}}{(n+2){3}^{n+1}}\cdot \frac{(n+1){3}^{n}}{{(x-2)}^{n}}| \\ & =\underset{n\to \infty }{\text{lim}}|\frac{(x-2)(n+1)}{3(n+2)}| \\ & =\frac{|x-2|}{3}.\end{array}\]
      The ratio \(\rho <1\) if \(|x-2|<3.\) Since \(|x-2|<3\) implies that \(-31\) if \(|x-2|>3.\) Therefore, the series diverges if \(x<-1\) or \(x>5.\) The ratio test is inconclusive if \(\rho =1.\) The ratio \(\rho =1\) if and only if \(x=-1\) or \(x=5.\) We need to test these values of x separately. For \(x=-1,\) the series is given by
      \[\sum _{n=0}^{\infty }\frac{{(-1)}^{n}}{n+1}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\text{\cdots }.\]
      Since this is the alternating harmonic series, it converges. Thus, the series converges at \(x=-1.\) For \(x=5,\) the series is given by
      \[\sum _{n=0}^{\infty }\frac{1}{n+1}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\text{\cdots }.\]
      This is the harmonic series, which is divergent. Therefore, the power series diverges at \(x=5.\) We conclude that the interval of convergence is \([-1,5)\) and the radius of convergence is \(R=3.\)

2. Find the interval and radius of convergence for the series \(\sum _{n=1}^{\infty }\frac{{x}^{n}}{\sqrt{n}}.\)

   Kusonyeza yankho

   The interval of convergence is \([-1,1).\) The radius of convergence is \(R=1.\)

3. Sketch a graph of \(f(x)=\frac{1}{1-x}\) and the graphs of the corresponding partial sums \({S}_{N}(x)=\sum _{n=0}^{N}{x}^{n}\) for \(N=2,4,6\) on the interval \((-1,1).\) Comment on the approximation \({S}_{N}\) as N increases.

   Kusonyeza yankho

   From the graph in you see that as N increases, \({S}_{N}\) becomes a better approximation for \(f(x)=\frac{1}{1-x}\) for x in the interval \((-1,1).\)

4. Sketch a graph of \(f(x)=\frac{1}{1-{x}^{2}}\) and the corresponding partial sums \({S}_{N}(x)=\sum _{n=0}^{N}{x}^{2n}\) for \(N=2,4,6\) on the interval \((-1,1).\)

   Kusonyeza yankho

   
   

5. Use a power series to represent each of the following functions \(f.\) Find the interval of convergence.

   1. \(f(x)=\frac{1}{1+{x}^{3}}\)
   2. \(f(x)=\frac{{x}^{2}}{4-{x}^{2}}\)
   Kusonyeza yankho

   1. You should recognize this function f as the sum of a geometric series, because
      \[\frac{1}{1+{x}^{3}}=\frac{1}{1-(\text{-}{x}^{3})}.\]
      Using the fact that, for \(|r|<1,\frac{a}{1-r}\) is the sum of the geometric series
      \[\sum _{n=0}^{\infty }a{r}^{n}=a+ar+a{r}^{2}+\text{\cdots },\]
      we see that, for \(|\text{-}{x}^{3}|<1,\)
      \[\begin{array}{ll}\frac{1}{1+{x}^{3}} & =\frac{1}{1-(\text{-}{x}^{3})} \\ & =\sum _{n=0}^{\infty }{(\text{-}{x}^{3})}^{n} \\ & =1-{x}^{3}+{x}^{6}-{x}^{9}+\text{\cdots }.\end{array}\]
      Since this series converges if and only if \(|\text{-}{x}^{3}|<1,\) the interval of convergence is \((-1,1),\) and we have
      \[\frac{1}{1+{x}^{3}}=1-{x}^{3}+{x}^{6}-{x}^{9}+\text{\cdots }\ \text{for}\ |x|<1.\]
   2. This function is not in the exact form of a sum of a geometric series. However, with a little algebraic manipulation, we can relate f to a geometric series. By factoring 4 out of the two terms in the denominator, we obtain
      \[\begin{array}{ll}\frac{{x}^{2}}{4-{x}^{2}} & =\frac{{x}^{2}}{4(1-\frac{{x}^{2}}{4})} \\ & =\frac{{x}^{2}}{4(1-{(\frac{x}{2})}^{2})}.\end{array}\]
      Therefore, we have
      \[\begin{array}{ll}\frac{{x}^{2}}{4-{x}^{2}} & =\frac{{x}^{2}}{4(1-{(\frac{x}{2})}^{2})} \\ & =\frac{\frac{{x}^{2}}{4}}{1-{(\frac{x}{2})}^{2}} \\ & =\sum _{n=0}^{\infty }\frac{{x}^{2}}{4}{(\frac{x}{2})}^{2n}.\end{array}\]
      The series converges as long as \(|{(\frac{x}{2})}^{2}|<1\) (note that when \(|{(\frac{x}{2})}^{2}|=1\) the series does not converge). Solving this inequality, we conclude that the interval of convergence is \((-2,2)\) and
      \[\begin{array}{ll}\frac{{x}^{2}}{4-{x}^{2}} & =\sum _{n=0}^{\infty }\frac{{x}^{2n+2}}{{4}^{n+1}} \\ & =\frac{{x}^{2}}{4}+\frac{{x}^{4}}{{4}^{2}}+\frac{{x}^{6}}{{4}^{3}}+\text{\cdots }\end{array}\]
      for \(|x|<2.\)

6. Represent the function \(f(x)=\frac{{x}^{3}}{2-x}\) using a power series and find the interval of convergence.

   Kusonyeza yankho

   \(\sum _{n=0}^{\infty }\frac{{x}^{n+3}}{{2}^{n+1}}\) with interval of convergence \((-2,2)\)

7. If \(\sum _{n=1}^{\infty }{a}_{n}{x}^{n}\) converges, then \({a}_{n}{x}^{n}\to 0\) as \(n\to \infty .\)

   Kusonyeza yankho

   True. If a series converges then its terms tend to zero.

8. \(\sum _{n=1}^{\infty }{a}_{n}{x}^{n}\) converges at \(x=0\) for any real numbers \({a}_{n}.\)

9. Given any sequence \({a}_{n},\) there is always some \(R>0,\) possibly very small, such that \(\sum _{n=1}^{\infty }{a}_{n}{x}^{n}\) converges on \((\text{-}R,R).\)

   Kusonyeza yankho

   False. It would imply that \({a}_{n}{x}^{n}\to 0\) for \(|x|

10. If \(\sum _{n=1}^{\infty }{a}_{n}{x}^{n}\) has radius of convergence \(R>0\) and if \(|{b}_{n}|\le |{a}_{n}|\) for all n, then the radius of convergence of \(\sum _{n=1}^{\infty }{b}_{n}{x}^{n}\) is greater than or equal to R.

11. Suppose that \(\sum _{n=0}^{\infty }{a}_{n}{(x-3)}^{n}\) converges at \(x=6.\) At which of the following points might the series diverge? Use the fact that if \(\sum {a}_{n}{(x-c)}^{n}\) converges at x, then it converges at any point closer to c than x.

    1. \(x=1\)
    2. \(x=2\)
    3. \(x=3\)
    4. \(x=0\)
    5. \(x=5.99\)
    6. \(x=0.000001\)
    Kusonyeza yankho

    It must converge on \((0,6]\) and hence at: a. \(x=1;\) b. \(x=2;\) c. \(x=3;\) d. \(x=0;\) e. \(x=5.99;\) and f. \(x=0.000001.\)

12. Suppose that \(\sum _{n=0}^{\infty }{a}_{n}{(x+1)}^{n}\) converges at \(x=-2.\) At which of the following points must the series also converge? Use the fact that if \(\sum {a}_{n}{(x-c)}^{n}\) converges at x, then it converges at any point closer to c than x.

    1. \(x=2\)
    2. \(x=-1\)
    3. \(x=-3\)
    4. \(x=0\)
    5. \(x=0.99\)
    6. \(x=0.000001\)

13. \(\sum _{n=0}^{\infty }{a}_{n}{2}^{n}{x}^{n}\)

    Kusonyeza yankho

    \(|\frac{{a}_{n+1}{2}^{n+1}{x}^{n+1}}{{a}_{n}{2}^{n}{x}^{n}}|=2|x||\frac{{a}_{n+1}}{{a}_{n}}|\to 2|x|\) so \(R=\frac{1}{2}\)

14. \(\sum _{n=0}^{\infty }\frac{{a}_{n}{x}^{n}}{{2}^{n}}\)

15. \(\sum _{n=0}^{\infty }\frac{{a}_{n}{\pi }^{n}{x}^{n}}{{e}^{n}}\)

    Kusonyeza yankho

    \(|\frac{{a}_{n+1}{(\frac{\pi }{e})}^{n+1}{x}^{n+1}}{{a}_{n}{(\frac{\pi }{e})}^{n}{x}^{n}}|=\frac{\pi |x|}{e}|\frac{{a}_{n+1}}{{a}_{n}}|\to \frac{\pi |x|}{e}\) so \(R=\frac{e}{\pi }\)

16. \(\sum _{n=0}^{\infty }\frac{{a}_{n}{(-1)}^{n}{x}^{n}}{{10}^{n}}\)

17. \(\sum _{n=0}^{\infty }{a}_{n}{(-1)}^{n}{x}^{2n}\)

    Kusonyeza yankho

    \(|\frac{{a}_{n+1}{(-1)}^{n+1}{x}^{2n+2}}{{a}_{n}{(-1)}^{n}{x}^{2n}}|=|{x}^{2}||\frac{{a}_{n+1}}{{a}_{n}}|\to |{x}^{2}|\) so \(R=1\)

18. \(\sum _{n=0}^{\infty }{a}_{n}{(-4)}^{n}{x}^{2n}\)

19. \(\sum _{n=1}^{\infty }\frac{{(2x)}^{n}}{n}\)

    Kusonyeza yankho

    \({a}_{n}=\frac{{2}^{n}}{n}\) so \(\frac{{a}_{n+1}x}{{a}_{n}}\to 2x.\) so \(R=\frac{1}{2}.\) When \(x=\frac{1}{2}\) the series is harmonic and diverges. When \(x=-\frac{1}{2}\) the series is alternating harmonic and converges. The interval of convergence is \(I=[-\frac{1}{2},\frac{1}{2}).\)

20. \(\sum _{n=1}^{\infty }{(-1)}^{n}\frac{{x}^{n}}{\sqrt{n}}\)

21. \(\sum _{n=1}^{\infty }\frac{n{x}^{n}}{{2}^{n}}\)

    Kusonyeza yankho

    \({a}_{n}=\frac{n}{{2}^{n}}\) so \(\frac{{a}_{n+1}x}{{a}_{n}}\to \frac{x}{2}\) so \(R=2.\) When \(x=\text{\pm }2\) the series diverges by the divergence test. The interval of convergence is \(I=(-2,2).\)

22. \(\sum _{n=1}^{\infty }\frac{n{x}^{n}}{{e}^{n}}\)

23. \(\sum _{n=1}^{\infty }\frac{{n}^{2}{x}^{n}}{{2}^{n}}\)

    Kusonyeza yankho

    \({a}_{n}=\frac{{n}^{2}}{{2}^{n}}\) so \(R=2.\) When \(x=\text{\pm }2\) the series diverges by the divergence test. The interval of convergence is \(I=(-2,2).\)

24. \(\sum _{k=1}^{\infty }\frac{{k}^{e}{x}^{k}}{{e}^{k}}\)

25. \(\sum _{k=1}^{\infty }\frac{{\pi }^{k}{x}^{k}}{{k}^{\pi }}\)

    Kusonyeza yankho

    \({a}_{k}=\frac{{\pi }^{k}}{{k}^{\pi }}\) so \(R=\frac{1}{\pi }.\) When \(x=\text{\pm }\frac{1}{\pi }\) the series is an absolutely convergent p-series. The interval of convergence is \(I=[-\frac{1}{\pi },\frac{1}{\pi }].\)

26. \(\sum _{n=1}^{\infty }\frac{{x}^{n}}{n\text{!}}\)

27. \(\sum _{n=1}^{\infty }\frac{{10}^{n}{x}^{n}}{n\text{!}}\)

    Kusonyeza yankho

    \({a}_{n}=\frac{{10}^{n}}{n\text{!}},\frac{{a}_{n+1}x}{{a}_{n}}=\frac{10x}{n+1}\to 0<1\) so the series converges for all x by the ratio test and \(I=(\text{-}\infty ,\infty ).\)

28. \(\sum _{n=1}^{\infty }{(-1)}^{n}\frac{{x}^{n}}{\text{ln}\ (2n)}\)

29. \(\sum _{k=1}^{\infty }\frac{{(k\text{!})}^{2}{x}^{k}}{(2k)\text{!}}\)

    Kusonyeza yankho

    \({a}_{k}=\frac{{(k\text{!})}^{2}}{(2k)\text{!}}\) so \(\frac{{a}_{k+1}}{{a}_{k}}=\frac{{(k+1)}^{2}}{(2k+2)(2k+1)}\to \frac{1}{4}\) so \(R=4\)

30. \(\sum _{n=1}^{\infty }\frac{(2n)\text{!}{x}^{n}}{{n}^{2n}}\)

31. \(\sum _{k=1}^{\infty }\frac{k\text{!}}{1\cdot 3\cdot 5\text{\cdots }(2k-1)}{x}^{k}\)

    Kusonyeza yankho

    \({a}_{k}=\frac{k\text{!}}{1\cdot 3\cdot 5\text{\cdots }(2k-1)}\) so \(\frac{{a}_{k+1}}{{a}_{k}}=\frac{k+1}{2k+1}\to \frac{1}{2}\) so \(R=2\)

32. \(\sum _{k=1}^{\infty }\frac{2\cdot 4\cdot 6\text{\cdots }2k}{(2k)\text{!}}{x}^{k}\)

33. \(\sum _{n=1}^{\infty }\frac{{x}^{n}}{(\begin{array}{l}2n \\ n\end{array})}\) where \((\begin{array}{l}n \\ k\end{array})=\frac{n\text{!}}{k\text{!}(n-k)\text{!}}\)

    Kusonyeza yankho

    \({a}_{n}=\frac{1}{(\begin{array}{l}2n \\ n\end{array})}\) so \(\frac{{a}_{n+1}}{{a}_{n}}=\frac{{((n+1)\text{!})}^{2}}{(2n+2)\text{!}}\ \frac{2n\text{!}}{{(n\text{!})}^{2}}=\frac{{(n+1)}^{2}}{(2n+2)(2n+1)}\to \frac{1}{4}\) so \(R=4\)

34. \(\sum _{n=1}^{\infty }{\text{sin}}^{2}n{x}^{n}\)

35. \(\sum _{n=1}^{\infty }\frac{{(n\text{!})}^{3}}{(3n)\text{!}}{x}^{n}\)

    Kusonyeza yankho

    \(\frac{{a}_{n+1}}{{a}_{n}}=\frac{{(n+1)}^{3}}{(3n+3)(3n+2)(3n+1)}\to \frac{1}{27}\) so \(R=27\)

36. \(\sum _{n=1}^{\infty }\frac{{2}^{3n}{(n\text{!})}^{3}}{(3n)\text{!}}{x}^{n}\)

37. \(\sum _{n=1}^{\infty }\frac{n\text{!}}{{n}^{n}}{x}^{n}\)

    Kusonyeza yankho

    \({a}_{n}=\frac{n\text{!}}{{n}^{n}}\) so \(\frac{{a}_{n+1}}{{a}_{n}}=\frac{(n+1)\text{!}}{n\text{!}}\ \frac{{n}^{n}}{{(n+1)}^{n+1}}={(\frac{n}{n+1})}^{n}\to \frac{1}{e}\) so \(R=e\)

38. \(\sum _{n=1}^{\infty }\frac{(2n)\text{!}}{{n}^{2n}}{x}^{n}\)

39. \(f(x)=\frac{1}{x};a=1\) (Hint: \(\frac{1}{x}=\frac{1}{1-(1-x)})\)

    Kusonyeza yankho

    \(f(x)=\sum _{n=0}^{\infty }{(1-x)}^{n}\) on \(I=(0,2)\)

40. \(f(x)=\frac{1}{1-{x}^{2}};a=0\)

Sankhani wanu

Parts of this page are adapted from OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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