My Calculus II teacher left us the task of finding a sequence that is conditionally convergent, and that if we multiply it by $(-1)^n$ it converges. I think that sequence could be  $\frac{sen(n)}{n}$. But first I have to prove that it is conditionally convergent, for that I will use the Dirichlet test. And one of his hypotheses is that the partial sums of $\Sigma_{n=1}^\infty \sin(n)$ are bounded , but I don't know how to do it. I would appreciate if  you do not use very complicated things in the proof, since I would also have to demonstrate them.
122145
So I'm trying to prove some series converges, and I'm trying to show it by using the Drichlet test.   So I need to prove that the series of partial sums $\Sigma_{k=1}^\infty \sin(ki)$ is bounded. I tried proving it by dividing and multiplying with $2\cos(\frac{i}{2})$ and then using the trigonometric identity $2\cos(\beta)\sin(\alpha)=sin(\alpha+\beta)-sin(\alpha-\beta)$, which creates a telescopic series - but I didn't manage to bound the result.  Any assistance would be great! Thanks in advance!
57722
Not sure how to approach this problem below. I tried assuming the conditional but that doesn't seem to get me anywhere.     If for a sequence of non-negative real numbers $\{a_n\}$, we have $$\sum{a_kb_k}$$ converges whenever $$\sum{b_k^2}$$ converges. then show that $$\sum{a_k^2}$$ converges
61698
This is a nice problem on series convergence that I recently stumbled upon. Given a non-negative sequence of real numbers $(a_n)$ such that $$\sum_{n=1}^\infty a_n &lt; \infty,$$ show that there exists a non-decreasing sequence of non-negative numbers $b_n$ such that $$b_n \to \infty \quad\text{ and } \quad \sum_{n=1}^\infty a_n b_n &lt; \infty.$$ In other words, for every convergent series with non-negative terms, there is another convergent series with "substantially larger" terms.  I have a solution (see below), but maybe someone else has a different, simpler, and/or more elegant solution.
112943
If $(a_n)$ is a decreasing sequence of strictly positive numbers and if $\sum{a_n}$ is convergent, show that $\lim{na_n}=0$  Since $(a_n)$ is decreasing and bounded below, by Monotonic Convergence Theorem, $(a_n)$ converges. So, there exists $N$ such that $|a_n-L|&lt; \epsilon$   Since $\sum{a_n}$ is convergent, by the Divergence test, we have $\lim_n{a_n}=0$, which means there exists $N$ such that $|a_n-0|&lt; \epsilon$  Then I get stuck at here.  I try to figure out the statement's meaning by inserting some examples, like $a_n=\dfrac{1}{n^2}$  Can anyone guide me on this question?
111427
Previous related .  After I negated the definition of a convergent sequence, I ended up with the following mathematical statement:   $$\exists\ \epsilon &gt; 0,\ \forall\ N \in \mathbb R\ \exists\ \mathbb N \ni n &gt; N : |x_n - l| \ge \epsilon$$  Is this correct? I'd like clarification...  Anyway, I'm now asked to use my negation of the definition of convergence to prove that $(a_n)=((-1)^nn)$ is divergent...  Am I right in assuming it's enough to prove that there exists at least one $\varepsilon$ such that $|(-1)^nn -\mathscr l|\ge\varepsilon$? Can I set $\mathscr l$ to be any real number I please?
64951
Let $(a_n)_{n\in\mathbb{N}}$ be a sequence such that lim inf $|a_n|=0$. Prove that there is a subsequence $(a_{n_{k}})_{k\in\mathbb{N}}$ such that $\sum_{k=1}^\infty a_{n_k}$ converges.  I'm thought about showing that you can make a subsequence that is smaller then $1/n^2$. Is this the right strategy ? Any hints ? Should I use the Cauchy criterion ?
117760
Consider the sequence of functions  $$ f_n(x) = \frac{x^2}{n} $$ for $x \in \mathbb{R}$  I have shown that $f_n(x)$ converges pointwise towards $0$ for $n \rightarrow \infty$ but I am not sure whether my attempt for showing that it does not converge uniformly towards $0$ for $n \rightarrow \infty$ is correct. Do you mind verifying?   By negation we have that $f_n(x)$ does not converge towards $0$ for $n \rightarrow \infty$ if   $$\exists \epsilon &gt; 0 \forall n \in \mathbb{N} \exists x \in \mathbb{R} \exists n \in \mathbb{N}: n \geq N \Rightarrow |\frac{x^2}{n}| \geq \epsilon$$  Let $\epsilon = 1$. Then for all $n \in \mathbb{N}$ we can find a $x \in \mathbb{R}$ such that for $n \geq N$ that $|x^2/n| \geq 1$. Can I then just pick $x = n^{1/2}$ so we have that $$ |x^2/n| = |(n^{1/2})^2/n| = |n/n| = 1 \geq 1 $$ which means that $f_n(x)$ does not converge uniformly towards $0$ for $n \rightarrow \infty$. Is this ok?
107313
Let $x_1=1$ and $$x_{n+1} = \frac12\left(x_n + \frac2{x_n}\right).$$ Prove or disprove $(x_n)$ is convergent and show the limit.   When I tried working on it I found the sequence was bounded by square root of 2 and it is was monotone. But apparently the sequence is not bounded by square root of two and is not monotone. But I have no idea why. Any help would be greatly appreciated! Thanks!
72368
Is the sequence $\Big(\dfrac {\tan n}{n}\Big) $ convergent ? If not convergent , is it properly divergent i.e. tends to   either $+\infty$ or $-\infty$ ? ( Owing to $\tan (n+1)= \dfrac {\tan n + \tan 1}{1- \tan1 \tan n}$ and the non-covergence of   $\Big (\tan n \Big)$ it is easy to see that if $\Big(\dfrac {\tan n}{n}\Big) $ is convergent then the limit must be  $0$ , but I can not proceed further . )
98761
This question arises because of a problem I was doing (Bartle 3rd edition, section 9.4 problem 3). It was like this. Given $a_n$ a decreasing sequence of positive numbers and suppose that $$\sum_{n=0}^{\infty}{a_n \sin{(nx)}}$$ Converge uniformly (It doesn't specify the domain, so I guess is for every x). Prove that $n a_n \to 0$. Clearly $\frac{1}{n}$ fits the description of $a_n$, and $n \frac{1}{n} \to 1 \neq 0$, so this would prove that there is a mistake in the problem if $\sum{\frac{\sin{(nx)}}{n}}$ converge uniformly for all $x$. So my question is if $\sum{\frac{\sin{(nx)}}{n}}$ converge uniformly for every $x$.  (I know that the series converge uniformly for every x in $[\delta, 2\pi - \delta]$, for $0 &lt; \delta &lt;2\pi$ by using the Dirichlet criterion.)
81277
I have this statement and I need to say if it is true or false:  Let $\{a_n\}$ be a real sequence.  $$\lim_{n\to +\infty} a_n = 0 \quad \implies \quad \sum_{n=1}^{\infty}(-1)^na_n \quad \text{converges}$$  I know, from the Leibniz criterion that:     If $a_n \to 0$, $a_n$ is decreasing and positive then $\sum_{n=1}^{\infty}(-1)^na_n$ converges   From this fact I believe that the statement is false but I couldn't come up with an infinitesimal sequence that isn't decreasing and for that reason is divergent. I tried some function with $sin(\frac{1}{n})$ without any luck. Any help would be very appreciated, thank you!
49841
Let $x_1=a&gt;0$ and $x_{n+1}=x_n+\frac{1}{x_n} \forall n\in \mathbb N$.   Check whether the following sequence converges or diverges.   When I was in UG my teacher used derivative test for monotonicity.  $f(x)=x+\frac{1}{x}, f'(x)=1-\frac{1}{x^2}&gt;0(x&gt;1).$ So, $f(x)$ is increasing. How to prove the sequence is monotonic? Differentiation is coming after the sequences and series. By AM-GM inequality sequence is bounded below. $x_{n+1}=x_n+\frac{1}{x_n}\ge 2\sqrt{x_n.\frac{1}{x_n}}=2  \forall n\in \mathbb N$. How can I judge whether the sequence bounded above or not? Please help me.
95433
Showing the sequence converges $a_{1}=\frac{1}{2}$,  $a_{n+1}=\frac{1}{2+a_{n}}$.  I already know that if $(a_{n})$ converges then it does to  $\sqrt{2}-1$.But i dont't know how to prove that this sequence cenverges.  EDIT I think that the subsequence $(a_{2n+1})$ is monotonic decreasing and the subsequence $(a_{2n})$ is monotonic increasing.
23237
Let $(a_n)_n$ be a sequence of non-negative real numbers.      Prove that $\sum\limits_{n=1}^{\infty}a_n$ converges if and only if the infinite product $\prod\limits_{n=1}^{\infty}(1+a_n)$ converges.   I don't necessarily need a full solution, just a clue on how to proceed would be appreciated.  Could any convergence test like the ratio or the root test help?
17808
Consider the series   $ \sum_{n=1}^\infty a_n $ where   $$ a_n:=\left(\exp\left(\frac{(-1)^n}{n}\right)-1\right).  $$   Show that it is convergent but not absolutely convergent.      I tried the alternating test but I'm not sure if $\{|a_n|\}$ is decreasing or not. And I don't see how other tests might help. Any ideas?    [Disclaimer: this question, asked on Jan 9 '14 at 16:01, is NOT a duplicate of the , which was asked much later on Jul 19 '15 at 20:32]
93622
I am trying to prove some theorem, and in the proof of it I have to check the fact that if the sequence $\{a_n\}_{n\in \mathbb{N}}$ converges, then $$\lim_{n\to \infty} n(a_n - a_{n-1}) = 0. $$ I think that this is true because I have tried with many different sequences and it holds, but I do not know how to prove it.  If this is not true, what is a counterexample for it?
62673
I have a question that says "Show that there is a bounded sequence $x_n$ which is not convergent but has the property that $x_n - x_{n+1} \to 0$ as $n \to 0$.  What does this mean? Do I need to come up with an example or does the problem actually want me to prove such proposition?  By the way, I see that this sequence looks like Cauchy because of $x_n - x_{n+1} \to 0$ as $n \to 0$, but it is obviously not.