Results 1 to 11 of 11

Math Help - Analysis, infinite series

  1. #1
    Newbie
    Joined
    Oct 2008
    Posts
    6

    Analysis, infinite series

    OK, questions is about divergent, conditionally convergent, or abs convergent.

    <br />
\sum\limits_{n = 1}^\infty {\frac{{\left( { - 1} \right)^{\left\lfloor {\sqrt n } \right\rfloor } }}{n}}<br />

    the first few terms are as follows,

    -1, -1/2, -1/3, 1/4, 1/5, 1/6, 1/7, 1/8, -1/9......

    I know I should take into the case of positive and negative terms by themselves. Using the epsilon/delta definition would also help. Thanks


    EDIT: Sorry about mixing the variables guys, been a long day
    Last edited by Maxmw22; October 27th 2008 at 04:34 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,966
    Thanks
    1785
    Awards
    1
    Quote Originally Posted by Maxmw22 View Post
    OK, questions is about divergent, conditionally convergent, or abs convergent.
    the series is n=1 to infinity of [(-1)^(sqrt(floor of n))]/n
    I think that there is a miss-type in that question.
    Note that for positive integers, the floor function is \left\lfloor n \right\rfloor  = n \Rightarrow \quad \sqrt {\left\lfloor n \right\rfloor }  = \sqrt n .
    From what you wrote “the first few terms are as follows, -1, -1/2, -1/3, 1/4, 1/5, 1/6, 1/7, 1/8, -1/9...”
    I think this is what you mean \sum\limits_{n = 1}^\infty  {\frac{{\left( { - 1} \right)^{\left\lfloor {\sqrt n } \right\rfloor } }}{n}} .
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2008
    Posts
    6
    Quote Originally Posted by Plato View Post
    I think that there is a miss-type in that question.
    Note that for positive integers, the floor function is \left\lfloor n \right\rfloor  = n \Rightarrow \quad \sqrt {\left\lfloor n \right\rfloor }  = \sqrt n .
    From what you wrote “the first few terms are as follows, -1, -1/2, -1/3, 1/4, 1/5, 1/6, 1/7, 1/8, -1/9...”
    I think this is what you mean \sum\limits_{n = 1}^\infty  {\frac{{\left( { - 1} \right)^{\left\lfloor {\sqrt n } \right\rfloor } }}{n}} .
    yes floor of the sqrt, not sqrt of the floor, seems like I am making all kinds of mistakes...haha

    but yes your expression is the one I am dealing with.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Oct 2008
    Posts
    6
    just bumping in hope of some help.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by Maxmw22 View Post
    OK, questions is about divergent, conditionally convergent, or abs convergent.

    <br />
\sum\limits_{n = 1}^\infty {\frac{{\left( { - 1} \right)^{\left\lfloor {\sqrt n } \right\rfloor } }}{n}}<br />

    the first few terms are as follows,

    -1, -1/2, -1/3, 1/4, 1/5, 1/6, 1/7, 1/8, -1/9......
    It converges. Note H_N=\sum_{n=1}^N\frac{1}{n}, and S_N= \sum\limits_{n = 1}^N {\frac{{\left( { - 1} \right)^{\left\lfloor {\sqrt n } \right\rfloor } }}{n}}.

    The sign changes when n is a square, so that we find, grouping the terms between two squares: S_{N^2-1}=\sum_{k=1}^{N-1} (-1)^k\left(\frac{1}{k^2}+\frac{1}{k^2+1}+\cdots+\fr  ac{1}{(k+1)^2-1}\right) = \sum_{k=1}^{N-1} (-1)^k s_k where s_k=\sum_{n=k^2+1}^{(k+1)^2-1}\frac{1}{n}= H_{(k+1)^2-1}-H_{k^2}.

    The idea is to prove that \sum_k (-1)^k s_k converges. There are different ways to do that:
    - using the expansion H_n=\ln n+\gamma+O_n\left(\frac{1}{n^2}\right) to find that (-1)^n s_n=2\frac{(-1)^n}{n}+O_n\left(\frac{1}{n^2}\right) is the sum of the general terms of two convergent series;
    - applying directly the alternating series theorem by showing that (s_n)_{n\geq 1} is decreasing and converges to 0. Neither is trivial. To show that (s_n)_n is decreasing, I found (if there's no mistake in my computation) that using comparison with an integral works: show that s_n is less than some integral ( s_k\leq\int_{k^2-1}^{(k+1)^2-1}\frac{dx}{x}), and s_{n+1} is greater than another one, which is greater than the previous one (usual comparison with the integral of a decreasing function)... This comparison allows as well to show that s_n converges to 0. [I let you try to fill in the details]

    Once you've shown that \sum_k (-1)^k s_k converges, you must deduce that the initial series converges. However, the difference between S_n and \sum_{k=1}^N (-1)^k s_k (where N^2 is the greatest square less than n) is less than s_{N+1}, which converges to 0. There may be mistakes with the indices here, but this is the main idea.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Oct 2008
    Posts
    6
    Quote Originally Posted by Laurent View Post
    It converges. Note H_N=\sum_{n=1}^N\frac{1}{n}, and S_N= \sum\limits_{n = 1}^N {\frac{{\left( { - 1} \right)^{\left\lfloor {\sqrt n } \right\rfloor } }}{n}}.

    The sign changes when n is a square, so that we find, grouping the terms between two squares: S_{N^2-1}=\sum_{k=1}^{N-1} (-1)^k\left(\frac{1}{k^2}+\frac{1}{k^2+1}+\cdots+\fr  ac{1}{(k+1)^2-1}\right) = \sum_{k=1}^{N-1} (-1)^k s_k where s_k=\sum_{n=k^2+1}^{(k+1)^2-1}\frac{1}{n}= H_{(k+1)^2-1}-H_{k^2}.

    The idea is to prove that \sum_k (-1)^k s_k converges. There are different ways to do that:
    - using the expansion H_n=\ln n+\gamma+O_n\left(\frac{1}{n^2}\right) to find that (-1)^n s_n=2\frac{(-1)^n}{n}+O_n\left(\frac{1}{n^2}\right) is the sum of the general terms of two convergent series;
    - applying directly the alternating series theorem by showing that (s_n)_{n\geq 1} is decreasing and converges to 0. Neither is trivial. To show that (s_n)_n is decreasing, I found (if there's no mistake in my computation) that using comparison with an integral works: show that s_n is less than some integral ( s_k\leq\int_{k^2-1}^{(k+1)^2-1}\frac{dx}{x}), and s_{n+1} is greater than another one, which is greater than the previous one (usual comparison with the integral of a decreasing function)... This comparison allows as well to show that s_n converges to 0. [I let you try to fill in the details]

    Once you've shown that \sum_k (-1)^k s_k converges, you must deduce that the initial series converges. However, the difference between S_n and \sum_{k=1}^N (-1)^k s_k (where N^2 is the greatest square less than n) is less than s_{N+1}, which converges to 0. There may be mistakes with the indices here, but this is the main idea.
    thanks, more than likely I will use the alternating series test, as it is the only one defined so far in my book. And of course if it converges it only conditionally converges as it would then resemble the harmonic series. I just need to find a way to rearrange the summands in order to make it converge, I thank you greatly.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by Laurent View Post
    s_k=\sum_{n=k^2}^{(k+1)^2-1}\frac{1}{n}.

    The idea is to prove that \sum_k (-1)^k s_k converges.
    Another way to prove this (the main step in Laurent's proof) is to show directly that the sequence s_k decreases to 0.

    It certainly converges to 0, because s_k < \sum_{n=k^2}^{(k+1)^2-1}\frac{1}{k^2} = \frac{2k+1}{k^2}\to0 as k\to\infty.

    To see that it decreases, we have

    s_k-s_{k+1} =  \sum_{n=0}^{2k}\left(\frac1{k^2+n}-\frac1{(k+1)^2+n}\right)- \frac1{(k+2)^2-2} - \frac1{(k+2)^2-1}

    (pairing off the terms in the sum for s_k with those of s_{k+1}, which leaves the two extra terms at the end of s_{k+1} to be subtracted off separately).

    This is equal to \sum_{n=0}^{2k}\frac{2k+1}{(k^2+n)((k+1)^2+n)} - \frac{2(k+2)^2-3}{((k+2)^2-2)((k+2)^2-1)}, which is greater than \frac{(2k+1)^2 - 2(k+2)^2+3}{((k+2)^2-2)((k+2)^2-1)} > 0.
    Last edited by Opalg; October 29th 2008 at 11:13 AM. Reason: silly mistake
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Aug 2008
    Posts
    903
    Based on Laurent's analysis above, here's my attempt to prove it converges. I know I can't hit very well but I have a strong arm. Personally I think this problem illustrates a deep and profound property of mathematics: Just how much can we stretch it and still have it converge? There's no clear boundary between convergent and divergent series; a fractal zoo separates them I believe but I digress. Here goes:

    \begin{aligned}\sum_{n=1}^{\infty}\frac{(-1)^{\lfloor\sqrt{n}\rfloor}}{n}&= -\left(1+1/2+1/3\right) &=s_1 \\<br />
&+\left(1/4+1/5+1/6+1/7+1/8\right)&=s_2 \\<br />
&-\left(1/9+1/10+\cdots+1/15\right)&=s_3 \\<br />
&+1/16+\cdots+1/24&=s_4 \\<br />
&\vdots \\<br />
&(-1)^k\left(\frac{1}{k^2}+\frac{1}{k^2+1}+\frac{1}{k  ^2+2}+\cdots+\frac{1}{(k+1)^2-1}\right)&=s_k<br />
\end{aligned}

    so that:

    \sum_{n=1}^{\infty}\frac{(-1)^{\lfloor\sqrt{n}\rfloor}}{n}=\sum_{k=1}^{\infty  } (-1)^k s_k

    This is now an alternating series and by inspection, it's obvious that:

    \left(\frac{1}{k^2}+\cdots+\frac{1}{(k+1)^2-1}\right)>\left(\frac{1}{(k+1)^2}+\cdots+\frac{1}{  (k+2)^2-1}\right)

    Therefore, s_{k+1}<s_{k}. Thus by the Alternating series test, if \lim_{k\to\infty} s_k\to 0 then the original series converges. But:

    \lim_{k\to \infty}\left\{\int_{k^2}^{(k+1)^2-1}\frac{dx}{x}<s_k<\int_{k^2}^{(k+1)^2-1}\frac{dx}{x-1}\right\}\to 0 and thus by the Squeeze Theorem, s_k\to 0. Therefore the original series converges.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by shawsend View Post
    This is now an alternating series and by inspection, it's obvious that:

    \left(\frac{1}{k^2}+\cdots+\frac{1}{(k+1)^2-1}\right)>\left(\frac{1}{(k+1)^2}+\cdots+\frac{1}{  (k+2)^2-1}\right)
    The fact is that this is not obvious, at least for me (there are more terms in the second sum than in the first one). This is the reason for my comparison with integrals (something like s_k\geq \int(\cdots)=\cdots\geq \cdots = \int(\cdots)\geq s_{k+1} with appropriate things in the dots), and this is what OpAlg proves in a different manner (by estimating the difference s_{k}-s_{k+1} directly).

    Therefore, s_{k+1}<s_{k}. Thus by the Alternating series test, if \lim_{k\to\infty} s_k\to 0 then the original series converges. But:

    \lim_{k\to \infty}\left\{\int_{k^2}^{(k+1)^2-1}\frac{dx}{x}<s_k<\int_{k^2}^{(k+1)^2-1}\frac{dx}{x-1}\right\}\to 0 and thus by the Squeeze Theorem, s_k\to 0.
    For this convergence of s_k to 0, I mentioned the comparison with the integral I had done previously (btw, note that 0\leq s_k hence the upper bound is enough), but OpAlg gives a much simpler argument, which consists in bounding each term of the sum defining s_k by the greatest one (i.e. the first one).
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member
    Joined
    Aug 2008
    Posts
    903
    Quote Originally Posted by Laurent View Post
    The fact is that this is not obvious, at least for me (there are more terms in the second sum than in the first one).
    Sorry. Dumb mistake for me.

    I need to study you and Opalg's analysis more.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member
    Joined
    Aug 2008
    Posts
    903
    Ok I got it now. Sorry for not looking at your post more carefully first Opalg. Honestly if I was in class with either of you I'd just sit in the back row and try my best to keep quiet as I'd have no hope of ever making the highest grade in anything the whole semester. I would of course expect our professor to assign quite challenging problems to both of you; the clowns in the back row, less so.
    Last edited by shawsend; October 29th 2008 at 01:03 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Infinite series (Analysis)
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 17th 2012, 03:09 AM
  2. Complex analysis, infinite series expansion of a function
    Posted in the Differential Geometry Forum
    Replies: 9
    Last Post: November 14th 2012, 01:08 PM
  3. Replies: 3
    Last Post: September 30th 2010, 06:17 PM
  4. Replies: 2
    Last Post: September 16th 2009, 08:56 AM
  5. Replies: 1
    Last Post: May 5th 2008, 10:44 PM

Search Tags


/mathhelpforum @mathhelpforum