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Prove that partial sums of $\sum_{n=1}^{\infty}{z^n}, z \in \mathbb{C}, |z|=1$ are bounded

Show $$\sum_{n=1}^N e^{i n\theta}.$$ is bounded for $ 0< \theta < 2 \pi$ and $ \forall N \in \{1,2,3,\ldots\}$

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Jmaff
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  • What did you try? – Did Jan 01 '13 at 21:23
  • First off, I think that we can consider sums like $\sum_{n=1} ^{N} a_{1} e^{in\theta}$ because of the way the sequence is. From there I looked at De Moivres theorem but don't know if it really helps – Jmaff Jan 01 '13 at 21:25
  • Not sure I got it right: you say you are able to prove the result when $a_n=a_1$ for every $n$? How? – Did Jan 01 '13 at 21:27
  • Right off the top of my head, I would first try to change the sum by summation by parts. Why? In order to take advantage of the decreasing nature of the series, by having instead a summation of products involving $a_{n-1}-a_n$. These differences will be positive and have a finite sum, which just might come in handy. – Harald Hanche-Olsen Jan 01 '13 at 21:28
  • No I meant that because the sequence is decreasing I suppose we are really more interested in seeing if $$\sum_{n=1}^N e^{i n\theta}.$$ is bounded. Since that may be false though I think I will edit the question because I actually asked this question as part of a larger question that uses Dirichlet's test and notice that I actually need to really see that $$\sum_{n=1}^N a_n e^{i n\theta}.$$ is bounded only. – Jmaff Jan 01 '13 at 21:28
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    The question should be clarified: For each such $\theta$, the sum is bounded (as a function of $N$). It is not the case that the expression is bounded as a function of both $\theta$ and $N$. – Hagen von Eitzen Jan 01 '13 at 21:33
  • ah sorry I left the $a_{n} up there. – Jmaff Jan 01 '13 at 21:34
  • The tag should be real-analysis, which I changed it to, but got changed back by @Jmaff. Complex analysis means analysis done on the complex plane, not simply that complex numbers are involved. – Calvin Lin Jan 01 '13 at 21:46
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    The question was changed in a major way after several people have offered answers. Some may consider that rude; I know I do. I deleted my answer, as it is no longer relevant. – Harald Hanche-Olsen Jan 01 '13 at 21:48
  • I'll change it, but it was only changed before since I didn't need that question answered and no effort to be "rude". Thanks for your help with the question, I appreciate it. – Jmaff Jan 01 '13 at 21:52
  • I'll make a new question if this occurs again; as I can see your point. thanks for bringing it up. – Jmaff Jan 01 '13 at 22:08
  • This is a complete mess now. The accepted answer is to a different question. – mrf Jan 01 '13 at 22:36
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    I cannot believe it: at some point, the OP mutilated the title and the question, pretty soon an answer addressing the modified version was posted and accepted (rather than any of the others, answering the original, more interesting, question), then the title and the question were reverted to basically their ante version, still by the OP. In all, this strange ballet took about 20 minutes. All rudeness aside (and this aspect does matter), a consequence is that the accepted answer does not answer the current question. – Did Jan 01 '13 at 22:37
  • moral of the story: post more carefully. The question that pertains to the accepted answer is now up^o^ – Jmaff Jan 01 '13 at 22:48

5 Answers5

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Hint: $$\sum_{n=1}^N e^{in\theta}\ =\frac{e^{i(N+1)\theta}-e^{i\theta}}{e^{i\theta}-1}$$ Clearly, the RHS is bounded

I answered this question , when the question was saying show that $\sum_{n=1}^N e^{in\theta}\ $ is bounded.

Amr
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  • @ Amr Is it true that $$\frac{e^{i(N+1)\theta}-e^{i\theta}}{e^{i\theta}-1}$$ is a decreasing sequence as a function of N? I don't see how for all N this implies that $| \sum_{n=1}^{N} e^{ in\theta}| \leq M$ for some constant M. – Jmaff Jan 01 '13 at 22:38
  • The question was edited. This answer is an answer to an older version of the question – Amr Jan 01 '13 at 22:39
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You can conclude it based on Abel partial summation (The result is termed as generalized alternating test or Dirichlet test). We will prove the generalized statement first.

Consider the sum $S_N = \displaystyle \sum_{n=1}^N a(n)b(n)$. Let $B(n) = \displaystyle \sum_{n=1}^N b(n)$. If $a(n) \downarrow 0$ and $B(n)$ is bounded, then the series $\displaystyle \sum_{n=1}^{\infty} a(n)b(n)$ converges absolutely.

First note that from Abel summation, we have that $$\sum_{n=1}^N a(n) b(n) = \sum_{n=1}^N a(n)(B(n)-B(n-1)) = \sum_{n=1}^{N} a(n) B(n) - \sum_{n=1}^Na(n)B(n-1)\\ = \sum_{n=1}^{N} a(n) B(n) - \sum_{n=0}^{N-1}^Na(n+1)B(n) = a(N) B(N) - a(1)B(0) + \sum_{n=1}^{N-1} B(n) (a(n)-a(n+1))$$ Now if $B(n)$ is bounded i.e. $\vert B(n) \vert \leq M$ and $a(n)$ is decreasing, then we have that $$\sum_{n=1}^{N-1} \left \vert B(n) \right \vert (a(n)-a(n+1)) \leq \sum_{n=1}^{N-1} M (a(n)-a(n+1))\\ = M (a(1) - a(N)) \leq Ma(1)$$ Hence, we have that $\displaystyle \sum_{n=1}^{N-1} \left \vert B(n) \right \vert (a(n)-a(n+1))$ converges and hence $$\displaystyle \sum_{n=1}^{N-1} B(n) (a(n)-a(n+1))$$ converges absolutely. Now since $$\sum_{n=1}^N a(n) b(n) = a(N) B(N) + \sum_{n=1}^{N-1} B(n) (a(n)-a(n+1))$$ we have that $\displaystyle \sum_{n=1}^N a(n)b(n)$ converges absolutely. In your case, $b(n) = \exp(in \theta)$. Hence, $$B(N) = \sum_{n=1}^N b(n) = \sum_{n=1}^N \exp(in \theta) = \exp(i\theta) \left(\dfrac{\exp(i N \theta)-1}{\exp(i \theta) - 1} \right)$$which is bounded for all $\theta \in (0, 2\pi)$. Hence, we have that $$\sum_{n=1}^N a(n) \exp(i n\theta)$$ converges.

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The Generalized Dirichlet Test states that if $$ \left|\sum_{k=1}^na_k\right|\le A\lt\infty\tag{1} $$ independent of $n$ and $$ \sum_{k=1}^\infty|b_k-b_{k+1}|=B\lt\infty\tag{2} $$ and $$ \lim_{k\to\infty}b_k=0\tag{3} $$ then $$ \sum_{k=1}^\infty a_kb_k\quad\text{converges and}\quad\left|\sum_{k=1}^\infty a_kb_k\right|\le AB\tag{4} $$

This follows using $$ A_n=\sum_{k=1}^na_k\tag{5} $$ where $A_0=0$, and by considering $$ \begin{align} \sum_{k=1}^na_kb_k &=\sum_{k=1}^n(A_k-A_{k-1})b_k\\ &=\sum_{k=1}^nA_kb_k-\sum_{k=0}^{n-1}A_kb_{k+1}\\ &=A_nb_n+\sum_{k=1}^{n-1}A_k(b_k-b_{k+1})\tag{6} \end{align} $$ $(1)$ says that $|A_n|\le A$, then $(2)$, $(3)$, and $(6)$ yield $(4)$.

Noting that $$ \begin{align} \left|\sum_{k=1}^ne^{i\theta}\right| &=\left|\frac{e^{i(n+1)\theta}-e^{i\theta}}{e^{i\theta}-1}\right|\\ &\le\frac2{|e^{i\theta}-1|} \end{align} $$ The Generalized Dirichlet's Test says that $$ \sum_{k=1}^\infty a_ke^{ik\theta} $$ converges as long as $\theta\ne0\pmod{2\pi}$ and $a_k$ satisfies $(2)$ and $(3)$ above.

robjohn
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Hint: The case $\theta = \pi$ has most probably been done as a theorem in your class. Try adapting that solution.

Calvin Lin
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Hint: Try the ratio test and alternating series test. You'll have to use the version where you use the $\limsup$ and $\liminf$.

Christopher A. Wong
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