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Determine whether the following series converge and whether they even converge absolutely:

  1. $\displaystyle\sum\limits_{k=2}^\infty\frac{(-1)^k}{(\log(k))^k}$
  2. $\displaystyle\sum\limits_{k=1}^\infty\left(\frac{1}{k!}-\frac{3}{(k+1)!}\right)$
  3. $\displaystyle\sum\limits_{k=1}^\infty\frac{2(k!)^2}{(2k)!}$
  4. $\displaystyle\sum\limits_{k=1}^\infty\frac{3^{2k-1}}{k^2+k}$
  5. $\displaystyle\sum\limits_{k=1}^\infty\left(\frac{2+(-1)^k}{4}\right)^k$
  6. $\displaystyle\sum\limits_{k=1}^\infty(-1)^k\frac{k}{(k+1)(k+2)}$

So I have been able to work through half of the series and I would like to know whether my current attempts are correct and then I would like to get some hints on how to solve the other series I intentionally left out by now.


My current solutions

  1. I know that I have to use the alternating series test and show that $1/(\log(k))^k$ is a null sequence, however I do not know whether it suffices to just mention the monotonicity of $\log$.

  2. $\displaystyle\sum\limits_{k=1}^\infty\left(\frac{1}{k!}-\frac{3}{(k+1)!}\right) = \sum\limits_{k=1}^\infty\left(\frac{k!(k-2)}{k!(k+1)!}\right) = \sum\limits_{k=1}^\infty\left(\frac{k-2}{(k+1)!}\right)$. If we apply the ratio test we get: $$\large\left|\frac{\frac{k-1}{(k+2)!}}{\frac{k-2}{(k+1)!}}\right|\normalsize=\frac{k+1}{k^2+4k}=\frac{k(1+1/k)}{k(k-4/k)}\longrightarrow 0<1,\text{ hence the series converges absolutely.}$$

  3. The ratio test yields $$\large\left|\frac{\frac{2((k+1)!)^2}{(2(k+1))!}}{\frac{2(k!)^2}{(2k)!}}\right| \normalsize = \frac{2((k+1)!)^2(2k)!}{(2(k+1))!2(k!)^2}=\frac{(k+1)^2(2k)!}{(2(k+1))!}=\frac{(k+1)^2}{(2k+2)(2k+1)}=\frac{k^2+2k+1}{4k^2+6k+2}=\frac{k^2(1+2/k+1/k^2)}{k^2(4+6/k+2/k^2)}\longrightarrow \frac{1}{4}<1,\text{ hence the series converges absolutely.}$$

  4. The ratio test yields $$\large\left|\frac{\frac{3^{2(k+1)-1}}{(k+1)^2+(k+1)}}{\frac{3^{2k-1}}{k^2+k}}\right|\normalsize = \frac{3^{2k+1}(k^2+k)}{3^{2k-1}((k+1)^2+(k+1))} = \frac{9k^2+9k}{k^2+3k+2}=\frac{k^2(9+9/k)}{k^2(1+3/k+2/k^2)}\longrightarrow 9>1,\text{ hence the series diverges.}$$

  5. I have no idea at all... a hint might help me out.

  6. Using the alternating series test, we have to show that $\frac{k}{(k+1)(k+2)}$ is a null sequence. This is easily done by $$\frac{k}{(k+1)(k+2)}=\frac{1}{k+3+2/k}\longrightarrow 0.$$ The series does not converge absolutely based on the comparison test where $$\sum\limits_{k=1}^\infty \frac{1}{k+3+2/k}\geq \sum\limits_{k=1}^\infty \frac{1}{k}.$$


Thanks for your time and reviewing my attempts.

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For the first, note that very soon $(\log k)^k$ is awfully big. For the second, both halves converge trivially. Need not manipulate so much. –  André Nicolas Nov 19 '12 at 15:43
    
This isn't true: $$\sum\limits_{k=1}^\infty \frac{1}{k+3+2/k}\geq \sum\limits_{k=1}^\infty \frac{1}{k}$$ Or rather, it isn't true that each term of the left side is $\geq$ each term of the right side. –  Thomas Andrews Nov 19 '12 at 15:54
    
@ThomasAndrews: Would it be correct to remove the sums and to just look at the sequences rather than the series? –  Christian Ivicevic Nov 19 '12 at 15:59
    
@ChristianIvicevic It's just not true that $$\frac{1}{k+3+2/k}\geq \frac{1}{k}$$, so the comparison test fails to show what you want. You can easily adjust it to show that your series is not absolutely convergent, but what you have here is wrong. –  Thomas Andrews Nov 19 '12 at 16:01
1  
@ChristianIvicevic Is that even true for $k=1$? Is $\frac{1}{6}\geq \frac{1}{2}$? –  Thomas Andrews Nov 19 '12 at 17:36

2 Answers 2

up vote 1 down vote accepted

Good work. One observation: Because you are skilled at computation, perhaps you start computing too quickly. Below are some comments on the problems.

$1.$ Very soon $\log k \gt 2$. After that point, the absolute value of the $k$-th term is $\lt \frac{1}{2^k}$. By comparison with the geometric series $\sum_{k=2}^\infty \frac{1}{2^k}$, our series converges absolutely, and hence converges.

Your observation that our series is an alternating series is correct. The terms do indeed go to $0$. In fact they go to $0$ very fast, fast enough, by a lot, to ensure absolute convergence.

$2.$ Before "simplifying," note that $\sum_1^\infty \frac{1}{n!}$ is a standard series that converges (absolutely). If you do not wish to assume that, do a Ratio Test on that. Similarly, $3\sum_{1}^\infty \frac{1}{(n+1)!}$ converges absolutely, so our series converges.

Note that this "splitting" strategy is not always appropriate.

$3, 4.$ Ratio Test is appropriate, and well executed.

$5.$ The $k$-th term is positive, and $\le \left(\frac{3}{4}\right)^k$, so by comparison with the geometric series $\sum_{k=1}^\infty \left(\frac{3}{4}\right)^k$, our series converges.

$6.$ The terms alternate in sign, and have limit $0$. If we can show that the terms (ultimately) go down steadily in absolute value, we will be able to conclude that our series is an alternating series, and therefore converges. One way to show that the terms are steadily decreasing in absolute value is to look at $\frac{k}{(k+1)(k+2)}-\frac{k+1}{(k+2)(k+3)}$. Bring to a common denominator. The numerator is positive, with the unimportant exception of the case $k=1$.

Alternately, one can use calculus to show that after a while $\frac{x}{(x+1)(x+2)}$ is decreasing.

The series does not converge absolutely. This is because, informally, in the long run the terms behave like $\frac{1}{k}$. This observation can be made formal: Look up the Limit Comparison Test.

However, we can do a formal proof in a simpler way. Note that $k+1\le 2k$ and $k+2\le 3k$. It follows that $\frac{k}{(k+1)(k+2)}\ge \frac{k}{(2k)(3k)}=\frac{1}{6}\cdot \frac{1}{k}$.

Since $\sum \frac{1}{k}$ diverges, so does $\sum \frac{1}{6}\cdot\frac{1}{k}$, and therefore so does $\sum \frac{1}{(k+1)(k+2)}$.

So our series converges, but not absolutely.

share|improve this answer
    
Concerning 5 I do not understand what you mean by saying the $k$-th term is positive - shouldn't that include something like "for every even k"? Furthermore does that mean the series does not converge absolutely? –  Christian Ivicevic Nov 19 '12 at 17:10
    
Look at $2+(-1)^k$. This is alternately $1$ and $3$, so it is always positive. Definitely the series converges absolutely. All the terms are positive. If you still doubt it, calculate the first $3$ or $4$ terms. –  André Nicolas Nov 19 '12 at 17:19
    
After a break I will work through your comments and rethink my ideas. Thanks for your effort. –  Christian Ivicevic Nov 19 '12 at 17:23

no.5 $\sum_{k=1}^{\infty}(\frac{2+(-1)^{2k}}{4})^{2k}+\sum_{k=1}^{\infty}(\frac{2+(-1)^{2k-1}}{4})^{2k-1}$

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