Is $x_n=\sin(1)+\sin(2)+...+\sin(n)=\sum_{i=1}^n \sin(i)$ a Cauchy sequence? By defenition $|x_{p+m}-x_p|=|sin(p+1)+...+sin(p+m)|=|cos(p+1/2)-cos(p+m+1/2)|/sin(1/2)$
88009
I received a task to find out whether the following series converges:  $$\sum_{x=1}^\infty\sin(x)$$  On first look it seems simple, but as I keep thinking about it, there's not a single lemma or criterion that I can use to tackle the problem.  D'alembert ? Doesn't work: The following is meaningless IMHO: $\lim\limits_{x \to \infty}{}\frac{\sin(x+1)}{\sin(x)}$  That series isn't monotonic... you can't understand the rules for when will a member of the series be negative or positive.  All I know is that $\sin(x)$ is blocked between (-1) and 1. Though it's easy to see that $ \sum\limits_{x=1}^\infty\lvert\sin(x) \rvert$ diverges.    May I use  for $\pi$ in order to construct  2 subseries:   One that shows that $\sin(x)$ converges to the limit 1 Another one that shows that $\sin(x)$ converges to the limit 0   And we know that a series can't converge to 2 different numbers, hence it diverges?
74043
Does the series $$ \sum_{n=1}^\infty \frac{\sin^2(n)}{n} $$   converge?   I've tried to apply some tests, and I don't know how to bound the general term, so I must have missed something. Thanks in advance.
59518
If I have a sequence {$a_n$} that has the property of $\lim(a_{n+1}-a_n)=0$, does that make it a Cauchy Sequence. I think it doesn't because $a_n = \sum_{k=1}^n \frac{1}{k}$ is a counter example.   However, by definition, there exists a $M$ such that if $n \geq M$ then $|a_{n+1}-a_n| &lt; \frac{\epsilon}{m-n}$  Hence, we have $|a_m - a_{m-1}|+.....+|a_{n+1}-a_n|&lt;|a_m -a_n| &lt;\epsilon$  This proof doesn't work because I cannot be sure I can find a fixed $M$ which might change according to n.  But I wonder if I have an additional condition that says $a_n$ is bounded, I think then the proof works and that I should be able to find a fixed $M$. However, I don't know how to justify this. Maybe I am wrong. Can someone kindly help me figure out this problem. Thanks
1428
I heard this question from a professor a couple years ago. I still think about it...     Does the sequence $(a_n)_{n\in \mathbb N}$ with $$a_n=\sqrt[n]{|\sin(n)|}$$ converges ( to $1$ ) ?   I believe this question is related to rational approximation.
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Given that $x_n$ and $y_n$ are Cauchy sequences in $\mathbb{R} $, prove that $x_n y_n$ is Cauchy without the use of the Cauchy theorem stating that Cauchy $\Rightarrow$ convergence.  Attempt: Without that condition on not been able to use the theorem, the question becomes trivial.  Instead:  For all $\epsilon &gt; 0$ there exists an $N \in \mathbb{N}$ such that for $n,m \geq N, -\frac{\epsilon}{2} \leq x_n - x_m \leq \frac{\epsilon}{2}$ and similar statment for $y_n$. Multiply the above by $y_n$ and the equivalent statement for $y_n$ by $x_m$.  Then add these together.  The result is: $$|x_ny_n - x_my_m| &lt; \frac{\epsilon}{2}(x_m + y_n)$$  I have proved in a previous question that $x_n + y_n$ is Cauchy so could I apply that here and say for $n,m \geq N$, $x_n + y_n$ is Cauchy and hence convergent so tends to a finite limit for $n,m \geq N$.  This would mean my upper bound is a multiple of $\epsilon$ and since $\epsilon$ is arbritarily small, so is this upper bound. Hence Cauchy.  I don't think this would warrant a full proof in any case since by multiplying by $x_m$ and $y_n$, I am assuming they are positive so as to not reverse the inequality signs.  Nonetheless, I would appreciate some feedback on what I have done.   Many thanks
84887
Let $t_n= \frac{1}{n} [1+\frac{1}{\sqrt{2}}+.....+\frac{1}{\sqrt{n}}]$, n=1,2...  Then I am asked whether series$ \sum t_n $ converge or diverge. Also whether sequence $ t_n $ converge to zero or not.  I tried About sequence $t_n$, by Cauchy's 1st theorem, that since 1/$\sqrt{n} $ converges to zero so does t_n. But the seriex part is doubtful to me.?
55965
Here is an exercise, on analysis which i am stuck.    How do I prove that if $F_{n}(x)=\sum\limits_{k=1}^{n} \frac{\sin{kx}}{k}$, then the sequence $\{F_{n}(x)\}$ is boundedly convergent on $\mathbb{R}$?
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At  I've read that being $\sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}$ by using  we get $\sin'(x)=\sum_{n=0}^{\infty}(2n+1) \frac{(-1)^n}{(2n+1)!}x^{2n}=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}=\cos(x)$ but the page  says that if $f(x)=\sum_{n=0}^{\infty}c_n(x-a)^n$ then inside the radius of convergence of the power series we have $f'(x)=\sum_{n=1}^{\infty}nc_n(x-a)^{n-1}$ so shouldn't it be $\sin'(x)=\sum_{n=1}^{\infty}(2n+1)\frac{(-1)^n}{(2n+1)!}x^{2n+1-1}=\sum_{n=1}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}$?  Now I understand that this last result has a missing term due to the index of the series being $1$ and not $0$ but being the $1$ mandatory because of the above theorem and since trying to shift index doesn't help (we get $\sin'(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{(2n+2)!}x^{2n+2}$) how can the result that ProofWiki shows be rigorously justified?   (I mean, I think we can't just say "oh, there's a missing term!" after having applied the theorem and magically shift the index of the series to obtain the desired result)
109251
I start my proof with supposing that I have a Cauchy sequence $x^\Bbb{R^n}_k$ in $\Bbb{R^n}$- that is, $x^\Bbb{R^n}_k=(x_1, x_2,...,x_N,...,x_m,..,x_n...)$ ,where $x_i=(x^i_1,x^i_2,x^i_3,...,x^i_n)$ e.g. $x^i_2$ denotes the second element of the $i$th entry in $x^\Bbb{R^n}_k$.  Since $x^\Bbb{R^n}_k$ is a Cauchy sequence for any $\epsilon&gt;0$: $\exists N \in\Bbb{N}$: $\forall m,n&gt;N$: $d(x_m,x_n)&lt;\epsilon$, where $d$ denotes the Euclidian metric in $\Bbb{R^n}$.   Some proofs I come across assumes/shows that the fact that  $x^\Bbb{R^n}_k$ is a Cauchy is extended to the elements of  $x^\Bbb{R^n}_k$ - that is $x_i=(x^i_1,x^i_2,x^i_3,...,x^i_n)$ is Cauchy $\forall i$ (e.g. ), and I do not understand how. Any clue or, of course, a full proof is appreciated.
113494
Does the series $\sum_{n=1}^{\infty}((1/n)-\sin(1/n))$ converge..? Can anyone please give me a simple proof..
92558
Theorem 3.22 of Rudin's Principles of Mathematical Analysis says:     The series $\sum a_{n}$ of (real or) complex numbers converges iff for every $\varepsilon &gt; 0$, there is an integer $N$ such that    $$ \left| \sum_{k=n}^m a_{k} \right| \leq \varepsilon $$    if $m\geq n \geq N.$    My doubt is: The condition is that the sequence of partial sums {$s_{n}$} is Cauchy, i.e for every $\varepsilon&gt;0$ there is an integer $N$ such that  $$ \left|s_{m}-s_{n} \right|&lt;\varepsilon $$  if  $m \geq n \geq N.$  But $\left|s_{m}-s_{n} \right|$ would be $\left|\sum_{k=n+1}^m a_{k}\right|$, isn't it?  How come the summation starts from $n$ in the theorem? How are the two summations equivalent?
80461
I thought about sequences, and I had an idea.     Let's define a sequence $(a_i)$ such that    $$a_0 = c \qquad\text{and}\qquad a_{n+1}=\sin a_n$$   where $c$ is arbitrary constant. Let's also define $$S_N := \sum_{i=0}^Na_i=c\;+\;\sin c\;+\;\sin\sin c\;+\;\cdots\;+\;\underbrace{\sin\sin\cdots\sin}_{N\text{ times}} c$$      Is $\lim_{N\to\infty} S_N$ convergent or divergent?   The problem is that I have no idea how I can, or if it is even possible to, determine $a_n$. Or if it is possible to show whether the sum converges.
122827
Question: Find the sum of the series: $$\lim_{n \to \infty}\frac{\sin1}{1}+\frac{\sin2}{2}+\frac{\sin3}{3}+...+\frac{\sin n}{n}$$  I have no clue how to find this. Obviously I can see the sum will be convergent as the denominator gets increasingly bigger while the numerator is bound between $1$ and $-1$.
97684
If $x=(x_n)$ is a sequence of complex numbers such that the series $\sum  x_ny_n$ is convergent for all $y=(y_n)\in{c_0}$. Then prove that $x\in{\ell^1}$. Can anyone tell me what is the meaning of $c_0$  in this question and how to proceed on proving it.
87254
A sequence $\{x_j\}$ is said to be fast-Cauchy if $\sum_1^\infty d(x_j,x_{j+1})&lt;+\infty$. Show that every Cauchy sequence has a fast-Cauchy subsequence.   **My attempt:**Argue by contradiction, suppose there is no such subsequence. Then for any subsequence $\{{x_j}_k\}$,  $\sum_1^\infty d({x_j}_k,{x_j}_{k+1})$ diverges. Vaguely recall from an analysis course, if a series diverges, it doesn't not satisfy the Cauchy criterion. That is to say, there exists $\epsilon&gt;0$ such that for all $k$ and all $p\ge1$, $d({x_j}_{k+1},{x_j}_{k+2})+...+d({x_j}_{k+p}+{x_j}_{k+p+1})\ge\epsilon$.  Then I intuitively think that this is a contradiction, since if the original series is Cauchy, then after a large $N$, since $d(x_m,x_n)$ has to be very small, so that their sum should be smaller than given $\epsilon$, which is a contradiction. But I don't know how I can write down the concrete proof?
117288
Consider the series expansion for the exponential function at x = 1:  $$a_n := \sum\limits_{i = 0}^n{\frac{1}{i!}}$$  I want to prove that this is a Cauchy sequence, using the remainder formula for the exponential function. (That's the term for it I'm familiar with, anyway.) It would be nice if you could comment on whether this proof is correct or if it could be simplified without using additional theorems or such.  For the Cauchy criteria, for each $\epsilon &gt; 0$ there must be a $N(\epsilon) \in N$ so that for all $m, n &gt; N(\epsilon)$, $ |a_m - a_n| &lt; \epsilon$.  So, take any $\epsilon &gt; 0$. Now, choose $N(\epsilon)$ so that:  $${\frac{1}{(N+1)!}} &lt; \frac{\epsilon}{2}.$$  (This is always possible because ${\frac{1}{n!}}$ goes to 0 when $n$ goes to infinity.)  Now, according to the partial sum formula, we got that:  $$\sum\limits_{i = 0}^{\infty}{\frac{1}{i!}} - \sum\limits_{i = 0}^{N(\epsilon)}{\frac{1}{i!}} = \sum\limits_{i = N(\epsilon)}^{\infty}{\frac{1}{i!}} &lt; \epsilon$$  So if we choose any $m, n &gt; N(\epsilon)$, and assume that $m &gt; n$ (because when $m = n$, the Cauchy criteria is obviously true), it follows that:  $$ \sum\limits_{i = 0}^{m}{\frac{1}{i!}} - \sum\limits_{i = 0}^{n}{\frac{1}{i!}} = \sum\limits_{i = n}^{m}{\frac{1}{i!}} &lt; \sum\limits_{i = N(\epsilon)}^{\infty}{\frac{1}{i!}} &lt; \epsilon$$  which proofs the statement.  Edit: the remainder formula I'm familiar with and that I've been using states that:  $$\sum\limits_{i = 0}^{\infty}{\frac{1}{i!}} - \sum\limits_{i = 0}^{n}{\frac{1}{i!}} &lt; 2 \frac{1}{(n+1)!} $$  if I recall correctly. Using this, can't we conclude that $\sum\limits_{i = N(\epsilon)}^{\infty}{\frac{1}{i!}} &lt; \epsilon $ ?
39045
Is the series  $$ \sum_{n=1}^\infty \frac{|\sin n|^n}n\tag{1}$$  convergent？  If one want to use Abel's test, is   $$ \sum_{n=1}^\infty |\sin n|^n\tag{2}$$  convergent？  Thank you very much