This is an exercise that was given to me, and I would like to emphasize that I am not looking for an answer, but rather a pointer or a hint in the right direction, since it appears I've reached a bit of a dead end. We are looking at sequences over the real numbers (at least that's my assumption, and that's where I've been looking).  I've come with various results to help me try such a sequence:   The ratio test has to be inconclusive, as does the root test.  Since $a_n$ converges to $0$, so does $a_n^3$ The sequences $|a_n|$ cannot be monotone decreasing, otherwise $\sum a_n^3$ will converge.   Other various things I've noticed:   It's very easy to find $a_n$ such that $a_n$ converges but $a_n^2$ diverges. Is it possible to use that fact? Let $u_n$ be such a sequence. I tried to construct a sum of two sequences so that when we take the third power, $u_n^2$ appears in the binomial expansion. That "naive" approach didn't get me too far however. We can make the following observation:  \begin{align} \sum a_n^3 &amp;= a_1^3 + a_2^3+a_3^3+a_4^3\cdots \\ &amp; = (a_1+a_2)((a_1-a_2)^2 + a_1a_2) + (a_3+a_4)((a_3-a_4)^2 + a_3a_4) +\cdots \end{align} This hasn't taken me anywhere.    Other things I've thought about:   I think a natural form for the sequence would be the product of two functions, $f$ and $g$, such that one converges to $0$ much faster than the other. By using the comparison test, we could try to require that $(fg)^3 &gt; k_n$ for all n for some sequence $k_n$ whose associated series is divergent. I think it is unlikely we could find such a product though, since if $fg$ plummets fast enough, then $(fg)^3$ will plummet even faster.   I've tried a few things for lack of better options, using exp, trig functions, polynomials, none of my tries have worked so far, almost always as a product of two functions.    If someone could suggest a possible step forward, I would be very grateful!
120838
Suppose the infinite series $\sum a_n$ converges ($a_n \in \mathbb{R}$ not all positive or negative).  (1) Is it true that $\sum a_n^2$ converges?  (2) Is it true that $\sum a_n^3$ converges?  For (1), the answer is no. I found the counterexample $a_n = (-1)^n / \sqrt{n}$ which converges by alternating series, but $\sum a_n^2 = \sum1/n $ is divergent.  I'm unsure about proving truth of (2) and can't seem to find a counterexample.  Thanks for help.
114562
Decide if the series converges and prove it using comparison test:   $\sum_{k=1}^{\infty}\frac{1}{2k+3}$   $$\frac{1}{2k+3} &lt;\frac{1}{2k}&lt;\frac{1}{k}$$  and since $\sum_{k=1}^{\infty}\frac{1}{k}$ diverges (we have defined that in our readings, so I can claim it here), the complete series will diverge, too.  Did I do everything correctly? What makes me feel a bit unsure is that there is no "$\leq$" sign anywhere and when I look up the comparison test on the internet, I don't see any "&lt;" or ">" in the definition.  Is it correct anyway?
4075
I have to prove the  by tomorrow and I am really unconfident about what I did:  $$\sum_{n=1}^\infty a_n\text{ converges } \iff \sum_{n=1}^\infty 2^n a_{2^n}\text{ converges}$$  I did the following:  Let $(b_n)$ be a sequence as follow: $b_{2^k+m}:=a_{2^k}$ with $k\in\mathbb N_0$ and $0\leq m&lt;2^k$.  It's $a_{n+1}\leq a_n$ and so $0\leq a_{n+p}\leq a_n$ for all $n,p\in\mathbb N$.  So $\sum\limits_{n=1}^\infty b_n$ converges by the majorizing series $\sum\limits_{n=1}^\infty a_n$. And it's $\sum\limits_{n=0}^\infty b_n=\sum\limits_{n=0}^\infty\sum\limits_{m=0}^{2^n-1}a_{2^{n+1}}=\sum\limits_{n=1}^\infty 2^{n-1}a_{2^n}$ so $\Rightarrow$ is done.  For $\Leftarrow$ consider $c_{2^k+m}:=a_{2^k}$ with $k\in\mathbb N_0$ and $0\leq m&lt;2^k$.  It's $|a_n|\leq c_n$ and $\sum\limits_{n=0}^\infty c_n=\sum\limits_{n=0}^\infty\sum\limits_{m=0}^{2^n-1}a_{2^{n}}=\sum\limits_{n=1}^\infty 2^{n}a_{2^n}$ and so $\sum\limits_{n=1}^\infty a_n$ converges by the majorizing series $\sum\limits_{n=0}^\infty c_n$.  Is this in form and content correct?
107818
I'm having a lot of trouble figuring this one out.  Determine if  $\sum_{i=2}^{\infty} \frac{(-1)^n}{n+(-1)^n}$ converges or diverges  Both ratio and root test are inconclusive and I'm at a loss. Can anyone help me?
107521
I am assuming my teacher wants us to conclude they converge by being bounded and monotone, since part (i) is to show that $a_n$ is monotone increasing and $b_n$ is monotone decreasing.  I am confused though because I have never worked with a sequence defined with... another sequence.
61842
Let $(a_n)$ be a sequence of positive terms and suppose that $\sum_\limits{n=0}^{\infty} a_n$ converges. Show that $\sum_\limits{n=0}^{\infty} a_n^2$ converges.   This is in the section on the Comparison Test so that must be what I'm supposed to use.  But I don't see how.  $(a_n)^2$ might be smaller or larger than $a_n$ depending on $a_n$.  And I can't use the Comparison Test with some other series because there's no info here about how fast $\sum a_n$ converges.  Hmm.  Any hints?
44461
I would appreciate your help!  How can we show that if a sequence of positive real numbers $a_n$ converges to $a\in\mathbb{R}$ with $a&gt;0$, then $(a_0 a_1\cdots a_n)^{\frac{1}{n}}$ converges to $a$.
75814
Assuming arithmetic operations on $\overline{\mathbb{C}}$ (that's the extended complex plane) are defined via arithmetic operations on the corresponding sequences, I need to give examples showing why $0\cdot \infty$, $\infty/\infty$, and $0/0$ are meaningless.  I also had to show the same thing for $\infty - \infty$, but that was easy: Consider the sequences $\{1,3,3,5,5,7,7,9,\cdots \}$ and $\{1,2,3,4,5,6,7,8,\cdots \}$, both of which belong to the class $\infty$ (i.e., the equivalence class of all sequences converging to $\infty$). Then, subtracting the second sequence from the first, term-by-term, we obtain the resulting sequence $\{0,1,0,1,0,1,\cdots\}$, which does not belong to any convergence class, since it has no limit.  However, I can't approach the three cases I'm asking on here about in the same way. I must admit, I'm a little confused as to what is being asked. For reference, it is problems 2-4 on p. 79 of Markushevich's Theory of Functions of a Complex Variable, Vol I.  If someone could furnish me with an example, at least for $0 \cdot \infty$, and maybe either $0/0$ or $\infty/\infty$, perhaps I could use that to help me figure out the other case on my own.  Thank you.
67522
Please help me solve this. Please try to give some details also.  Let $a_n=\sum_{k=1}^n  \frac{n}{n^2+k}\ $for $ n \in\mathbb{N}$ then the sequence $\{a_n\}$ is (choose the correct option):     Convergent   Bounded but not convergent   Diverges to infinity   Neither bounded nor diverges to infinity     My thoughts : I tried doing it by testing if it is monotonic with help of $a_{n+1} -a_n$ and checking if its +ve or -ve but, the the terms dont simplyfy by cancelling each other...
75491
Convergent, absolutely convergent or divergent:   $$\sum_{k=1}^{\infty}\frac{2^{k}\cdot k!}{k^{k}}$$   I have used ratio test because we got a fraction here and I think I did alright till the end:  $$\lim_{k\rightarrow \infty}\frac{\frac{2^{k+1}\cdot (k+1)!}{(k+1)^{k+1}}}{\frac{2^{k}\cdot k!}{k^{k}}} = \lim_{k\rightarrow \infty}\frac{2^{k+1}\cdot (k+1)! \cdot k^{k}}{(k+1)^{k+1}\cdot 2^{k}\cdot k!} = \lim_{k\rightarrow \infty}\frac{2^{k}\cdot 2^1\cdot k! \cdot (k+1)\cdot k^{k}}{(k+1)^{k}\cdot (k+1)\cdot 2^{k}\cdot k!} = \lim_{k\rightarrow \infty}\frac{2k^{k}}{(k+1)^{k}}$$ $$=2\lim_{k\rightarrow \infty} \left( \frac{k}{k+1}\right)^{k}$$  Now I don't know (without a calculator) to what this would converge / diverge to... In the exam we are not allowed to use a calculator... So what to do?  The denominator will be greater than the enumerator by 1, so dividing each other we got something $&lt;1$. We take exponent $k$ which is $\geq 1$, so we will end up with $&lt;1$ again. Multiply this with 2 we get something $&lt; 1$ but $&gt; 0$ and thus the series is convergent...?  I hope I have described it well? Would you give me full points to this task? :D  Edit: I haven't described it well in the end. See the accepted answer and its comments! Thanks a lot to everyone - from every question I ask here, I always learn new things :-)
108692
This is the series $\sum_{k=1}^{\infty} \frac{7^{k} + k}{k! + 1}$. I began with the ratio test, but after I set up the ratio to take the limit of, I couldn't simplify the limit, since nothing would cancel out. I believe it converges, but I have no idea how to solve further after setting up that ratio test limit.  If someone has another method, that would be fine too. We are not allowed to use Comparison or Limit Comparison Tests.
29669
If each term $a_k$ is either 0 or 1, how many sequences are there $a_1, a_2, \cdots$ , $a_n$ such that    $$a_1 \leq a_2 \geq a_3 \leq a_4 \cdots?$$   If we let $f_n$ be the number of ways for a n-term sequence, we can get a recursive formula.   Suppose $a_1=1$. Then $a_2=1$ and $a_3=0,1$ which is actually $f_{n-2}$.  Suppose $a_1=0$. If $a_2=1$, we get $f_{n-2}$ by similar reasoning. If $a_2=0$, $a_3=0$ so the cycle repeats. Thus the # of ways in this case is $f_{n-2}+f_{n-4}+f_{n-6}+\cdots$.  So the recursive formula will be $$f_n=2f_{n-2}+f_{n-4}+f_{n-6}+\cdots$$ and $f_1=2, f_2=3$. How do I get a closed form, or did I find the wrong recursive relation?
56344
Let $\{a_n\}$ be defined as follows: $$a_1 = 2, a_{n+1}=\dfrac{1}{3-a_n}$$ $n \geq 1$.   Does $\{a_n\}$  converge?    Using the monotone convergence thereom, if the sequence is both bounded below and decreasing, $\forall n\in\mathbb{N}, n\geq 1$, then we can say the sequence converges.  I'm having trouble proving that the sequence is bounded below by induction.    Claim: The sequence is bounded below by $0$.  Base Case:  $n=1$  $a_1 = 2 \geq 0$, so this holds.  Induction Step  Suppose that $a_k \geq 0$, for some $k \in\mathbb{N}$. (IH)  Prove $a_k \geq 0\to a_{k+1} \geq 0$  I have some difficulty here.  $a_{k}=\dfrac{1}{3-a_{k-1}} \geq 0$  $a_{k+1}=\dfrac{1}{3-a_{k}}$, but we know $a_k \geq 0$, but we don't know how much bigger. It could be that $a_k = 4$, and then I have $a_{k+1}=\dfrac{1}{3-4}$, which is negative, and does not confirm with my claim of being bounded below by $0$. Have I gone somewhere wrong in my induction?
68767
Let $(a_n)_{n≥0}$ and $(b_n)_{n≥0}$ be two sequences of a nomed algebra such that $\sum{\| a_n\|}$ and $\sum{\| b_n\|}$ converge, and$ $$$\forall n, \  a_n, b_n \neq 0$$  Show that $(\forall  k \in \mathbb N^*,  \ \sum_{n=0}^{\infty}{a_n}^k = \sum_{n=0}^{\infty}{b_n}^k) \implies \exists\ \sigma \in{\mathfrak S}_{\mathbb N} \ ,\ (a_n) = (b_{\sigma(n)})$  The vector-space should be a finite-dimensional $\mathbb R$-vector-space.   I do not see how to deal with the hypothesis. If you have ideas/hints...  It would be relevant to deal with complex sequences even if it is less general. But I don't think one can generalize using real and complex cases.
121001
Prove convergence\divergence of the series: $$\sum_{n=1}^{\infty}\dfrac {1\cdot 3\cdots (2n-1)} {2\cdot 4\cdots (2n)}$$  Here is what I have at the moment:  Method I  My first way uses a result that is related to  that we'll denote by $W_{n}$.  Also, we  may denote $\dfrac {1\cdot 3\cdots (2n-1)} {2\cdot 4\cdots (2n)}$ by $P_{n}$. Having noted these and taking a large value of $n$ we get: $$(P_{n})^2 =\frac{1}{W_{n} \cdot (2n+1)}\approx\frac{2}{\pi}\cdot \frac{1}{2n+1}$$ $$P_{n}\approx \sqrt {\frac{2}{\pi}} \cdot \frac{1}{\sqrt{2n+1}}$$   Further we have that: $$\lim_{n\to\infty}\sqrt {\frac{2}{\pi}} \cdot \frac{n}{\sqrt{2n+1}} \le \sum_{n=1}^{\infty} P_{n}$$  that obviously shows us that the series diverges.  Method II  The second way is to resort to the powerful  and firstly proceed with the ratio test: $$\lim_{n\to\infty} \frac{P_{n+1}}{P_{n}}=\frac{2n+1}{2n+2}=1$$ and according to the result, the ratio test is inconclusive.  Now, we apply Kummer's test and get: $$\lim_{n\to\infty} \frac{P_{n}}{P_{n+1}}n-(n+1)=\lim_{n\to\infty} -\frac{n+1}{2n+1}=-\frac{1}{2} \le 0$$ Since  $$\sum_{n=1}^{\infty} \frac{1}{n} \longrightarrow \infty$$ our series diverges and we're done.  On the site I've also found   with answers that can be applied for my question.  Since I've already have some answers for my question you may regard it as a recreational one and if you have a nice proof to share I'd be glad to receive it. I like this question very much and want to make up a collection with nice proofs for it. Thanks.
8737
Suppose $a_n \ge 0$ for all natural numbers $\Bbb{N}$ and $\sum{a_n}$ converges. Is it necessarily true that $\sum{(a_n)^2}$ converges?  I think yes.  Clearly as $a_n \ge 0$ it follows $(a_1+a_2+\dots+a_n)^2 \ge a_1^2+a_2^2+\dots+a_n^2 \ge 0$ and so $(\sum_{n=1}^{\infty}{a_n})^2 \ge \sum_{n=1}^{\infty}{a_n^2} \ge 0$.  As we are told $\sum{a_n}$ converges to say $L$ where $L$ is a real number it follows $L^2 \ge \sum_{n=1}^{\infty}{a_n^2} \ge 0$ and so therefore $\sum{(a_n)^2}$ converges.  Is this correct?
7590
I am trying to check whether or not the sequence $$a_{n} =\left\{\frac{n^n}{n!}\right\}_{n=1}^{\infty}$$ is bounded, convergent and ultimately monotonic (there exists an $N$ such that for all $n\geq N$ the sequence is monotonically increasing or decreasing). However, I'm having a lot of trouble finding a solution that sufficiently satisfies me.  My best argument so far is as follows,  $$a_{n} = \frac{n\cdot n\cdot n\cdot \ldots\cdot n}{n(n-1)(n-2)(n-3)\dots(2)(1)} = \frac{n}{n}\cdot \frac{n}{n-1}\cdot  \ldots \cdot \frac{n}2\cdot n$$  so $\lim a_{n}\rightarrow \infty$ since $n&lt;a_{n}$ for all $n&gt;1$. Since the sequence is divergent, it follows that the function must be ultimately monotonic.  This feels a little dubious to me, I feel like I can form a much better argument than that, or at the very least a more elegant one. I've tried to assume $\{a_{n}\}$ approaches some limit $L$ so there exists some $N$ such that  $|a_{n} - L| &lt; \epsilon$ whenever $n&gt;N$ and derive a contradiction, but this approach got me nowhere.  Finally, I've also tried to use the fact that $\frac{a_{n+1}}{a_n}\rightarrow e$ to help me, but I couldn't find an argument where that fact would be useful.