Review: Convergence of Series

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Test the following series for convergence or divergence:

  1. \displaystyle{\sum_{n=1}^\infty \frac{1}{n+3^n}}
  2. \displaystyle{\sum_{n=1}^\infty \frac{(2n+1)^n}{n^{2n}}}
  3. \displaystyle{\sum_{n=1}^\infty (-1)^n \frac{n}{n+2}}
  4. \displaystyle{\sum_{n=1}^\infty (-1)^n \frac{n}{n^2+2}}
  5. \displaystyle{\sum_{n=1}^\infty \frac{n^2 2^{n-1}}{(-5)^n}}
  6. \displaystyle{\sum_{n=1}^\infty \frac{1}{2n+1}}
  7. \displaystyle{\sum_{n=2}^\infty \frac{1}{n \sqrt{\ln n}}}
  8. \displaystyle{\sum_{n=1}^\infty \frac{2^n n!}{(n+2)!}}
  9. \displaystyle{\sum_{n=1}^\infty n^2 e^{-n}}
  10. \displaystyle{\sum_{n=1}^\infty n^2 e^{-n^3}}
  11. \displaystyle{\sum_{n=2}^\infty \frac{(-1)^{n+1}}{n \ln n}}
  12. \displaystyle{\sum_{n=1}^\infty \sin n}
  13. \displaystyle{\sum_{n=1}^\infty \frac{3^n n^2}{n!}}
  14. \displaystyle{\sum_{n=1}^\infty \frac{\sin 2n}{1+2^n}}
  15. \displaystyle{\sum_{n=0}^\infty \frac{n!}{2 \cdot 5 \cdot 8 \cdot \dotsb \cdot (3n+2)}}
  16. \displaystyle{\sum_{n=1}^\infty \frac{n^2+1}{n^3+1}}
  17. \displaystyle{\sum_{n=1}^\infty (-1)^n 2^{1/n}}
  18. \displaystyle{\sum_{n=2}^\infty \frac{(-1)^{n-1}}{\sqrt{n} - 1}}
  19. \displaystyle{\sum_{n=1}^\infty (-1)^n \frac{\ln n}{\sqrt{n}}}
  1. \displaystyle{\sum_{n=1}^\infty \frac{n+5}{5^n}}
  2. \displaystyle{\sum_{n=1}^\infty \frac{(-2)^{2n}}{n^n}}
  3. \displaystyle{\sum_{n=1}^\infty \frac{\sqrt{n^2-1}}{n^3+2n^2+5}}
  4. \displaystyle{\sum_{n=1}^\infty \tan (1/n)}
  5. \displaystyle{\sum_{n=1}^\infty n \sin(1/n)}
  6. \displaystyle{\sum_{n=1}^\infty \frac{n!}{e^{n^2}}}
  7. \displaystyle{\sum_{n=1}^\infty \frac{n^2+1}{5^n}}
  8. \displaystyle{\sum_{n=1}^\infty \frac{n \ln n}{(n+1)^3}}
  9. \displaystyle{\sum_{n=1}^\infty \frac{e^{1/n}}{n^2}}
  10. \displaystyle{\sum_{n=1}^\infty \frac{(-1)^n}{\cosh n}}
  11. \displaystyle{\sum_{n=1}^\infty (-1)^n \frac{\sqrt{n}}{n+5}}
  12. \displaystyle{\sum_{n=1}^\infty \frac{5^n}{3^n+4^n}}
  13. \displaystyle{\sum_{n=1}^\infty \frac{(n!)^n}{n^{4n}}}
  14. \displaystyle{\sum_{n=1}^\infty \frac{\sin(1/n)}{\sqrt{n}}}
  15. \displaystyle{\sum_{n=1}^\infty \frac{1}{n+n\cos^2 n}}
  16. \displaystyle{\sum_{n=1}^\infty \bigg( \frac{n}{n+1} \bigg)^{n^2}}
  17. \displaystyle{\sum_{n=2}^\infty \frac{1}{\big( \ln n \big)^{\ln n}}}
  18. \displaystyle{\sum_{n=1}^\infty \big( 2^{1/n} -1 \big)^n}
  19. \displaystyle{\sum_{n=1}^\infty \big( 2^{1/n} -1 \big)}
  1. Davis
    November 3, 2011 at 2:40 pm
    Reply

    On number five, I did by the ratio test. However, for a hypothetical example similar, let’s say the ratio test gives a limit of less than one, but the alternating series test shows the series increasing. Which test do I listen to, which takes precedence? Or is this even possible?

    • Francisco Blanco-Silva
      November 3, 2011 at 8:50 pm
      Reply

      That is an excellent question! Alternating series test takes precedence over ratio/root.

  2. Bushra
    November 5, 2011 at 3:11 pm
    Reply

    for the alternating series test, if both conditions are null, then does that automatically mean that the series is divergent, OR simply that it is not convergent and the test for diveregence needs to be done?

    • Francisco Blanco-Silva
      November 5, 2011 at 7:58 pm
      Reply

      As soon as one of the conditions is not satisfied, the test cannot be applied. You need to use ratio or root instead.

  3. Bushra
    November 5, 2011 at 3:59 pm
    Reply

    also for alternating series problems do we need to specify whther they or conditionally or absolutely convergent

  4. Davis
    November 7, 2011 at 8:20 pm
    Reply

    For number 21, you do the root test, but does the series also have to satisfy the alternating series test?

    • Francisco Blanco-Silva
      November 7, 2011 at 8:42 pm
      Reply

      21 is not an alternating series.

      • Davis
        November 7, 2011 at 8:45 pm
        Reply

        Whoops, I see that now. But if it was and it was decreasing, you could do the ratio test correct?

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