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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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marked as duplicate by Did, Fabian, Jonas Teuwen, ncmathsadist, QiL Jan 2 '13 at 0:04

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

What did you try? – Did Jan 1 '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 1 '13 at 21:25
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 1 '13 at 21:33
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 1 '13 at 21:48
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 1 '13 at 22:37
up vote 4 down vote accepted

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.

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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 1 '13 at 22:38
The question was edited. This answer is an answer to an older version of the question – Amr Jan 1 '13 at 22:39

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.

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

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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 $\liminf$.

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