Sequences

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 sequences, find an explicit formula for the \(n\text{th}\) term of the sequence.

   1. \(-\frac{1}{2},\frac{2}{3},-\frac{3}{4},\frac{4}{5},-\frac{5}{6}\text{,\ldots }\)
   2. \(\frac{3}{4},\frac{9}{7},\frac{27}{10},\frac{81}{13},\frac{243}{16}\text{,\ldots }\)
   జవాబు వెల్లడి చేయండి

   1. First, note that the sequence is alternating from negative to positive. The odd terms in the sequence are negative, and the even terms are positive. Therefore, the \(n\text{th}\) term includes a factor of \({(-1)}^{n}.\) Next, consider the sequence of numerators \(\{1,2,3\text{,\ldots }\}\) and the sequence of denominators \(\{2,3,4\text{,\ldots }\}.\) We can see that both of these sequences are arithmetic sequences. The \(n\text{th}\) term in the sequence of numerators is \(n,\) and the \(n\text{th}\) term in the sequence of denominators is \(n+1.\) Therefore, the sequence can be described by the explicit formula
      \[{a}_{n}=\frac{{(-1)}^{n}n}{n+1}.\]
   2. The sequence of numerators \(3,9,27,81,243\text{,\ldots }\) is a geometric sequence. The numerator of the \(n\text{th}\) term is \({3}^{n}\) The sequence of denominators \(4,7,10,13,16\text{,\ldots }\) is an arithmetic sequence. The denominator of the \(n\text{th}\) term is \(4+3(n-1)=3n+1.\) Therefore, we can describe the sequence by the explicit formula \({a}_{n}=\frac{{3}^{n}}{3n+1}.\)

2. Find an explicit formula for the \(n\text{th}\) term of the sequence \(\{\frac{1}{5},-\frac{1}{7},\frac{1}{9},-\frac{1}{11}\text{,\ldots }\}.\)

   జవాబు వెల్లడి చేయండి

   \({a}_{n}=\frac{{(-1)}^{n+1}}{3+2n}\)

3. For each of the following recursively defined sequences, find an explicit formula for the sequence.

   1. \({a}_{1}=2,\) \({a}_{n}=-3{a}_{n-1}\) for \(n\ge 2\)
   2. \({a}_{1}=\frac{1}{2},\) \({a}_{n}={a}_{n-1}+{(\frac{1}{2})}^{n}\) for \(n\ge 2\)
   జవాబు వెల్లడి చేయండి

   1. Writing out the first few terms, we have
      \[\begin{array}{l}{a}_{1}=2 \\ {a}_{2}=-3{a}_{1}=-3(2) \\ {a}_{3}=-3{a}_{2}={(-3)}^{2}2 \\ {a}_{4}=-3{a}_{3}={(-3)}^{3}2.\end{array}\]
      In general,
      \[{a}_{n}=2{(-3)}^{n-1}.\]
   2. Write out the first few terms:
      \[\begin{array}{l} \\ \\ {a}_{1}=\frac{1}{2} \\ {a}_{2}={a}_{1}+{(\frac{1}{2})}^{2}=\frac{1}{2}+\frac{1}{4}=\frac{3}{4} \\ {a}_{3}={a}_{2}+{(\frac{1}{2})}^{3}=\frac{3}{4}+\frac{1}{8}=\frac{7}{8} \\ {a}_{4}={a}_{3}+{(\frac{1}{2})}^{4}=\frac{7}{8}+\frac{1}{16}=\frac{15}{16}.\end{array}\]
      From this pattern, we derive the explicit formula
      \[{a}_{n}=\frac{{2}^{n}-1}{{2}^{n}}=1-\frac{1}{{2}^{n}}.\]

4. Find an explicit formula for the sequence defined recursively such that \({a}_{1}=-4\) and \({a}_{n}={a}_{n-1}+6.\)

   జవాబు వెల్లడి చేయండి

   \({a}_{n}=6n-10\)

5. For each of the following sequences, determine whether or not the sequence converges. If it converges, find its limit.

   1. \(\{5-\frac{3}{{n}^{2}}\}\)
   2. \(\{\frac{3{n}^{4}-7{n}^{2}+5}{6-4{n}^{4}}\}\)
   3. \(\{\frac{{2}^{n}}{{n}^{2}}\}\)
   4. \(\{{(1+\frac{4}{n})}^{n}\}\)
   జవాబు వెల్లడి చేయండి

   1. We know that \(1\text{/}n\to 0.\) Using this fact, we conclude that
      \[\underset{n\to \infty }{\text{lim}}\frac{1}{{n}^{2}}=\underset{n\to \infty }{\text{lim}}(\frac{1}{n}).\underset{n\to \infty }{\text{lim}}(\frac{1}{n})=0.\]
      Therefore,
      \[\underset{n\to \infty }{\text{lim}}(5-\frac{3}{{n}^{2}})=\underset{n\to \infty }{\text{lim}}5-3\underset{n\to \infty }{\text{lim}}\frac{1}{{n}^{2}}=5-3\cdot 0=5.\]
      The sequence converges and its limit is \(5.\)
   2. By factoring \({n}^{4}\) out of the numerator and denominator and using the limit laws above, we have
      \[\begin{array}{ll}\underset{n\to \infty }{\text{lim}}\frac{3{n}^{4}-7{n}^{2}+5}{6-4{n}^{4}} & =\underset{n\to \infty }{\text{lim}}\frac{3-\frac{7}{{n}^{2}}+\frac{5}{{n}^{4}}}{\frac{6}{{n}^{4}}-4} \\ & =\frac{\underset{n\to \infty }{\text{lim}}(3-\frac{7}{{n}^{2}}+\frac{5}{{n}^{4}})}{\underset{n\to \infty }{\text{lim}}(\frac{6}{{n}^{4}}-4)} \\ & =\frac{(\underset{n\to \infty }{\text{lim}}(3)\text{-}\underset{n\to \infty }{\text{lim}}\frac{7}{{n}^{2}}+\underset{n\to \infty }{\text{lim}}\frac{5}{{n}^{4}})}{(\underset{n\to \infty }{\text{lim}}\frac{6}{{n}^{4}}-\underset{n\to \infty }{\text{lim}}(4))} \\ & =\frac{(\underset{n\to \infty }{\text{lim}}(3)\text{-}7\cdot \underset{n\to \infty }{\text{lim}}\frac{1}{{n}^{2}}+5\cdot \underset{n\to \infty }{\text{lim}}\frac{1}{{n}^{4}})}{(6\cdot \underset{n\to \infty }{\text{lim}}\frac{1}{{n}^{4}}-\underset{n\to \infty }{\text{lim}}(4))} \\ & =\frac{3-7\cdot 0+5\cdot 0}{6\cdot 0-4}=-\frac{3}{4}.\end{array}\]
      The sequence converges and its limit is \(-3\text{/}4.\)
   3. Consider the related function \(f(x)={2}^{x}\text{/}{x}^{2}\) defined on all real numbers \(x>0.\) Since \({2}^{x}\to \infty\) and \({x}^{2}\to \infty\) as \(x\to \infty ,\) apply L’Hôpital’s rule and write
      \[\begin{array}{lllll}\underset{x\to \infty }{\text{lim}}\frac{{2}^{x}}{{x}^{2}} & =\underset{x\to \infty }{\text{lim}}\frac{{2}^{x}\text{ln}\ 2}{2x} & & & \text{Take the derivatives of the numerator and denominator.} \\ & =\underset{x\to \infty }{\text{lim}}\frac{{2}^{x}{(\text{ln}\ 2)}^{2}}{2} & & & \text{Take the derivatives again.} \\ & =\infty . & & & \end{array}\]
      We conclude that the sequence diverges.
   4. Consider the function \(f(x)={(1+\frac{4}{x})}^{x}\) defined on all real numbers \(x>0.\) This function has the indeterminate form \({1}^{\infty }\) as \(x\to \infty .\) Let
      \[y=\underset{x\to \infty }{\text{lim}}{(1+\frac{4}{x})}^{x}.\]
      Now taking the natural logarithm of both sides of the equation, we obtain
      \[\text{ln}(y)=\text{ln}[\underset{x\to \infty }{\text{lim}}{(1+\frac{4}{x})}^{x}].\]
      Since the function \(f(x)=\text{ln}\ x\) is continuous on its domain, we can interchange the limit and the natural logarithm. Therefore,
      \[\text{ln}(y)=\underset{x\to \infty }{\text{lim}}[\text{ln}{(1+\frac{4}{x})}^{x}].\]
      Using properties of logarithms, we write
      \[\underset{x\to \infty }{\text{lim}}[\text{ln}{(1+\frac{4}{x})}^{x}]=\underset{x\to \infty }{\text{lim}}x\ \text{ln}(1+\frac{4}{x}).\]
      Since the right-hand side of this equation has the indeterminate form \(\infty \cdot 0,\) rewrite it as a fraction to apply L’Hôpital’s rule. Write
      \[\underset{x\to \infty }{\text{lim}}x\ \text{ln}(1+\frac{4}{x})=\underset{x\to \infty }{\text{lim}}\frac{\text{ln}(1+4\text{/}x)}{1\text{/}x}.\]
      Since the right-hand side is now in the indeterminate form \(0\text{/}0,\) we are able to apply L’Hôpital’s rule. We conclude that
      \[\underset{x\to \infty }{\text{lim}}\frac{\text{ln}(1+4\text{/}x)}{1\text{/}x}=\underset{x\to \infty }{\text{lim}}\frac{4}{1+4\text{/}x}=4.\]
      Therefore, \(\text{ln}(y)=4\) and \(y={e}^{4}.\) Therefore, since \(\underset{x\to \infty }{\text{lim}}{(1+\frac{4}{x})}^{x}={e}^{4},\) we can conclude that the sequence \(\{{(1+\frac{4}{n})}^{n}\}\) converges to \({e}^{4}.\)

6. Consider the sequence \(\{(5{n}^{2}+1)\text{/}{e}^{n}\}.\) Determine whether or not the sequence converges. If it converges, find its limit.

   జవాబు వెల్లడి చేయండి

   The sequence converges, and its limit is \(0.\)

7. Determine whether the sequence \(\{\text{cos}(3\text{/}{n}^{2})\}\) converges. If it converges, find its limit.

   జవాబు వెల్లడి చేయండి

   Since the sequence \(\{3\text{/}{n}^{2}\}\) converges to \(0\) and \(\text{cos}\ x\) is continuous at \(x=0,\) we can conclude that the sequence \(\{\text{cos}(3\text{/}{n}^{2})\}\) converges and

   \[\underset{n\to \infty }{\text{lim}}\text{cos}(\frac{3}{{n}^{2}})=\text{cos}(0)=1.\]

8. Determine if the sequence \(\{\sqrt{\frac{2n+1}{3n+5}}\}\) converges. If it converges, find its limit.

   జవాబు వెల్లడి చేయండి

   The sequence converges, and its limit is \(\sqrt{2\text{/}3}.\)

9. Use the Squeeze Theorem to find the limit of each of the following sequences.

   1. \(\{\frac{\text{cos}\ n}{{n}^{2}}\}\)
   2. \(\{{(-\frac{1}{2})}^{n}\}\)
   జవాబు వెల్లడి చేయండి

   1. Since \(-1\le \text{cos}\ n\le 1\) for all integers \(n,\) we have
      \[-\frac{1}{{n}^{2}}\le \frac{\text{cos}\ n}{{n}^{2}}\le \frac{1}{{n}^{2}}.\]
      Since \(-1\text{/}{n}^{2}\to 0\) and \(1\text{/}{n}^{2}\to 0,\) we conclude that \(\text{cos}\ n\text{/}{n}^{2}\to 0\) as well.
   2. Since
      \[-\frac{1}{{2}^{n}}\le {(-\frac{1}{2})}^{n}\le \frac{1}{{2}^{n}}\]
      for all positive integers \(n,\) \(-1\text{/}{2}^{n}\to 0\) and \(1\text{/}{2}^{n}\to 0,\) we can conclude that \({(-1\text{/}2)}^{n}\to 0.\)

10. Find \(\underset{n\to \infty }{\text{lim}}\frac{2n-\text{sin}\ n}{n}.\)

    జవాబు వెల్లడి చేయండి

    \(2\)

11. For each of the following sequences, use the Monotone Convergence Theorem to show the sequence converges and find its limit.

    1. \(\{\frac{{4}^{n}}{n\text{!}}\}\)
    2. \(\{{a}_{n}\}\) defined recursively such that
       \[{a}_{1}=2\ \text{and}\ {a}_{n+1}=\frac{{a}_{n}}{2}+\frac{1}{2{a}_{n}}\ \text{for all}\ n\ge 2.\]
    జవాబు వెల్లడి చేయండి

    1. Writing out the first few terms, we see that
       \[\{\frac{{4}^{n}}{n\text{!}}\}=\{4,8,\frac{32}{3},\frac{32}{3},\frac{128}{15}\text{,\ldots }\}.\]
       At first, the terms increase. However, after the third term, the terms decrease. In fact, the terms decrease for all \(n\ge 3.\) We can show this as follows.
       \[{a}_{n+1}=\frac{{4}^{n+1}}{(n+1)\text{!}}=\frac{4}{n+1}\cdot \frac{{4}^{n}}{n\text{!}}=\frac{4}{n+1}\cdot {a}_{n}\le {a}_{n}\ if\ n\ge 3.\]
       Therefore, the sequence is decreasing for all \(n\ge 3.\) Further, the sequence is bounded below by \(0\) because \({4}^{n}\text{/}n\text{!}\ge 0\) for all positive integers \(n.\) Therefore, by the Monotone Convergence Theorem, the sequence converges.
       To find the limit, we use the fact that the sequence converges and let \(L=\underset{n\to \infty }{\text{lim}}{a}_{n}.\) Now note this important observation. Consider \(\underset{n\to \infty }{\text{lim}}{a}_{n+1}.\) Since
       \[\{{a}_{n+1}\}=\{{a}_{2,}{a}_{3},{a}_{4}\text{,\ldots }\},\] the only difference between the sequences \(\{{a}_{n+1}\}\) and \(\{{a}_{n}\}\) is that \(\{{a}_{n+1}\}\) omits the first term. Since a finite number of terms does not affect the convergence of a sequence,
       \[\underset{n\to \infty }{\text{lim}}{a}_{n+1}=\underset{n\to \infty }{\text{lim}}{a}_{n}=L.\]
       Combining this fact with the equation
       \[{a}_{n+1}=\frac{4}{n+1}{a}_{n}\]
       and taking the limit of both sides of the equation
       \[\underset{n\to \infty }{\text{lim}}{a}_{n+1}=\underset{n\to \infty }{\text{lim}}\frac{4}{n+1}{a}_{n},\]
       we can conclude that
       \[L=0\cdot L=0.\]
    2. Writing out the first several terms,
       \[\{2,\frac{5}{4},\frac{41}{40},\frac{3281}{3280}\text{,\ldots }\}.\]
       we can conjecture that the sequence is decreasing and bounded below by \(1.\) To show that the sequence is bounded below by \(1,\) we can show that
       \[\frac{{a}_{n}}{2}+\frac{1}{2{a}_{n}}\ge 1.\]
       To show this, first rewrite
       \[\frac{{a}_{n}}{2}+\frac{1}{2{a}_{n}}=\frac{{a}_{n}^{2}+1}{2{a}_{n}}.\]
       Since \({a}_{1}>0\) and \({a}_{2}\) is defined as a sum of positive terms, \({a}_{2}>0.\) Similarly, all terms \({a}_{n}>0.\) Therefore,
       \[\frac{{a}_{n}^{2}+1}{2{a}_{n}}\ge 1\]
       if and only if
       \[{a}_{n}^{2}+1\ge 2{a}_{n}.\]
       Rewriting the inequality \({a}_{n}^{2}+1\ge 2{a}_{n}\) as \({a}_{n}^{2}-2{a}_{n}+1\ge 0,\) and using the fact that
       \[{a}_{n}^{2}-2{a}_{n}+1={({a}_{n}-1)}^{2}\ge 0\]
       because the square of any real number is nonnegative, we can conclude that
       \[\frac{{a}_{n}}{2}+\frac{1}{2{a}_{n}}\ge 1.\]
       To show that the sequence is decreasing, we must show that \({a}_{n+1}\le {a}_{n}\) for all \(n\ge 1.\) Since \(1\le {a}_{n}^{2},\) it follows that
       \[{a}_{n}^{2}+1\le 2{a}_{n}^{2}.\]
       Dividing both sides by \(2{a}_{n},\) we obtain
       \[\frac{{a}_{n}}{2}+\frac{1}{2{a}_{n}}\le {a}_{n}.\]
       Using the definition of \({a}_{n+1},\) we conclude that
       \[{a}_{n+1}=\frac{{a}_{n}}{2}+\frac{1}{2{a}_{n}}\le {a}_{n}.\]
       Since \(\{{a}_{n}\}\) is bounded below and decreasing, by the Monotone Convergence Theorem, it converges.
       To find the limit, let \(L=\underset{n\to \infty }{\text{lim}}{a}_{n}.\) Then using the recurrence relation and the fact that \(\underset{n\to \infty }{\text{lim}}{a}_{n}=\underset{n\to \infty }{\text{lim}}{a}_{n+1},\) we have
       \[\underset{n\to \infty }{\text{lim}}{a}_{n+1}=\underset{n\to \infty }{\text{lim}}(\frac{{a}_{n}}{2}+\frac{1}{2{a}_{n}}),\]
       and therefore
       \[L=\frac{L}{2}+\frac{1}{2L}.\]
       Multiplying both sides of this equation by \(2L,\) we arrive at the equation
       \[2{L}^{2}={L}^{2}+1.\]
       Solving this equation for \(L,\) we conclude that \({L}^{2}=1,\) which implies \(L=\text{\pm }1.\) Since all the terms are positive, the limit \(L=1.\)

12. Consider the sequence \(\{{a}_{n}\}\) defined recursively such that \({a}_{1}=1,\) \({a}_{n}={a}_{n-1}\text{/}2.\) Use the Monotone Convergence Theorem to show that this sequence converges and find its limit.

    జవాబు వెల్లడి చేయండి

    \(0.\)

13. \({a}_{n}=1+{(-1)}^{n}\) for \(n\ge 1\)

    జవాబు వెల్లడి చేయండి

    \({a}_{n}=0\) if \(n\) is odd and \({a}_{n}=2\) if \(n\) is even

14. \({a}_{n}={n}^{2}-1\) for \(n\ge 1\)

15. \({a}_{1}=1\) and \({a}_{n}={a}_{n-1}+n\) for \(n\ge 2\)

    జవాబు వెల్లడి చేయండి

    \(\{{a}_{n}\}=\{1,3,6,10,15,21\text{,\ldots }\}\)

16. \({a}_{1}=1,\) \({a}_{2}=1\) and \({a}_{n+2}={a}_{n}+{a}_{n+1}\) for \(n\ge 1\)

17. Find an explicit formula for \({a}_{n}\) where \({a}_{1}=1\) and \({a}_{n}={a}_{n-1}+n\) for \(n\ge 2.\)

    జవాబు వెల్లడి చేయండి

    \({a}_{n}=\frac{n(n+1)}{2}\)

18. Find a formula \({a}_{n}\) for the \(n\text{th}\) term of the arithmetic sequence whose first term is \({a}_{1}=1\) such that \({a}_{n+1}-{a}_{n}=17\) for \(n\ge 1.\)

19. Find a formula \({a}_{n}\) for the \(n\text{th}\) term of the arithmetic sequence whose first term is \({a}_{1}=-3\) such that \({a}_{n+1}-{a}_{n}=4\) for \(n\ge 1.\)

    జవాబు వెల్లడి చేయండి

    \({a}_{n}=4n-7\)

20. Find a formula \({a}_{n}\) for the \(n\text{th}\) term of the geometric sequence whose first term is \({a}_{1}=1\) such that \(\frac{{a}_{n+1}}{{a}_{n}}=10\) for \(n\ge 1.\)

21. Find a formula \({a}_{n}\) for the \(n\text{th}\) term of the geometric sequence whose first term is \({a}_{1}=3\) such that \(\frac{{a}_{n+1}}{{a}_{n}}=1\text{/}10\) for \(n\ge 1.\)

    జవాబు వెల్లడి చేయండి

    \({a}_{n}={3.10}^{1-n}={30.10}^{\text{-}n}\)

22. Find an explicit formula for the \(n\text{th}\) term of the sequence whose first several terms are \(\{0,3,8,15,24,35,48,63,80,99\text{,\ldots }\}.\) (Hint: First add one to each term.)

23. Find an explicit formula for the \(n\text{th}\) term of the sequence satisfying \({a}_{1}=0\) and \({a}_{n}=2{a}_{n-1}+1\) for \(n\ge 2.\)

    జవాబు వెల్లడి చేయండి

    \({a}_{n}={2}^{n-1}-1\)

24. \(\{1,0,-1,0,1,0,-1,0\text{,\ldots }\}\) (Hint: Find where \(\text{sin}\ x\) takes these values)

25. \(\{1,\text{-}1\text{/}3,1\text{/}5,\text{-}1\text{/}7\text{,\ldots }\}\)

    జవాబు వెల్లడి చేయండి

    \({a}_{n}=\frac{{(-1)}^{n-1}}{2n-1}\)

26. \({a}_{1}=1\) and \({a}_{n+1}=\text{-}{a}_{n}\) for \(n\ge 1\)

27. \({a}_{1}=2\) and \({a}_{n+1}=2{a}_{n}\) for \(n\ge 1\)

    జవాబు వెల్లడి చేయండి

    \(f(n)={2}^{n}\)

28. \({a}_{1}=1\) and \({a}_{n+1}=(n+1){a}_{n}\) for \(n\ge 1\)

29. \({a}_{1}=2\) and \({a}_{n+1}=(n+1){a}_{n}\text{/}2\) for \(n\ge 1\)

    జవాబు వెల్లడి చేయండి

    \(f(n)=n\text{!}\text{/}{2}^{n-2}\)

30. \({a}_{1}=1\) and \({a}_{n+1}={a}_{n}\text{/}{2}^{n}\) for \(n\ge 1\)

31. [T] \({a}_{1}=1,\) \({a}_{2}=2,\) and for \(n\ge 2,\) \({a}_{n}=\frac{1}{2}({a}_{n-1}+{a}_{n-2});\) \(N=30\)

    జవాబు వెల్లడి చేయండి

    Terms oscillate above and below \(5\text{/}3\) and appear to converge to \(5\text{/}3.\)
    

32. [T] \({a}_{1}=1,\) \({a}_{2}=2,\) \({a}_{3}=3\) and for \(n\ge 4,\) \({a}_{n}=\frac{1}{3}({a}_{n-1}+{a}_{n-2}+{a}_{n-3}),\) \(N=30\)

33. [T] \({a}_{1}=1,\) \({a}_{2}=2,\) and for \(n\ge 3,\) \({a}_{n}=\sqrt{{a}_{n-1}{a}_{n-2}};\) \(N=30\)

    జవాబు వెల్లడి చేయండి

    Terms oscillate above and below \(y\approx 1.57...\) and appear to converge to a limit.
    

34. [T] \({a}_{1}=1,\) \({a}_{2}=2,\) \({a}_{3}=3,\) and for \(n\ge 4,\) \({a}_{n}=\sqrt{{a}_{n-1}{a}_{n-2}{a}_{n-3}};\) \(N=30\)

35. \(\underset{n\to \infty }{\text{lim}}(3{a}_{n}-4{b}_{n})\)

    జవాబు వెల్లడి చేయండి

    \(7\)

36. \(\underset{n\to \infty }{\text{lim}}(\frac{1}{2}{b}_{n}-\frac{1}{2}{a}_{n})\)

37. \(\underset{n\to \infty }{\text{lim}}\frac{{a}_{n}+{b}_{n}}{{a}_{n}-{b}_{n}}\)

    జవాబు వెల్లడి చేయండి

    \(0\)

38. \(\underset{n\to \infty }{\text{lim}}\frac{{a}_{n}-{b}_{n}}{{a}_{n}+{b}_{n}}\)

39. \(\frac{{n}^{2}}{{2}^{n}}\)

    జవాబు వెల్లడి చేయండి

    \(0\)

40. \(\frac{{(n-1)}^{2}}{{(n+1)}^{2}}\)

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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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