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$$\sum n^{1/5}\sin\frac1n=\sum n^{-4/5}\frac{\sin\frac1n}{\frac1n}=\infty$$ since $\frac{\sin\frac1n}{\frac1n}\to1$ and $\sum n^{-4/5}=\infty$.

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$f(x)=x^2$ is not uniformly continuos over $\mathbb C$ and maps Cauchy sequences into Cauchy sequences.

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Since $f$ is uniformly continuous, by definition, given $\epsilon > 0$, there exists some $\delta > 0$ (only depending on $\epsilon$) such that $$\vert f(x) - f(y) \vert < \epsilon$$ whenever $x, y\in D$ satisfies $\vert x - y \vert < \delta$. Let $(x_n)$ be a Cauchy sequence in $D$, then for $\delta > 0$ (as above), there exists some ...

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No, it's not. You need to start with one Cauchy sequence, say $(x_n)$. Then you need to show that $\bigl(f(x_n)\bigr)$ is a Cauchy sequence. To do this, first fix an $\epsilon>0$. You now need to show that there is an $N$ so that $\bigl|f(x_n)-f(x_m)\bigr|<\epsilon$ whenever $n,m\ge N$. To do this, first use the uniform continuity of $f$ to find ...

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Here it is another formal proof, Given $\epsilon>0$ there exists $A(\epsilon)> \frac{1}{\epsilon}$ such that for $a\geq A(\epsilon)$ we have; $\displaystyle{|a_{(n)}-0|=\frac{a}{4^a}< \frac{2^a}{4^a}= \frac{1}{2^a} < \frac{1}{a}<\frac{1}{A(\epsilon)}< \frac{1}{\frac{1}{\epsilon}}}= \epsilon$ Which concludes the proof.

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$\sum_i \sum_n l(J_n^i)$ is interpreted as $\lim_{j\rightarrow\infty}\sum_{i=1}^j\sum_n l(J_n^i)$. Now, you should prove that there's a $\sigma$ such that $\sum_k l(J_{\sigma_1(k)}^{\sigma_2(k)})$ converges. To do that, you take any $\sigma$ and bound the finite sums with $\sum_i \sum_n l(J_n^i)$. As the terms are positive, $\sum_k ... 0 It's false:$x_n=(-1)^n=y_n$,$n_i=2i$,$m_i=2i+1$, then$x_{n_i}=1$and$y_{m_i}=-1$. If you take$\{l_i\}_i$such that$x_{l_i}\rightarrow 1$then exists$i_0$such that$x_{l_i}=1\quad\forall i\ge i_0\Rightarrow l_i$is even$\forall i\ge i_0\Rightarrow y_{l_i}=1\quad\forall i\ge i_0\Rightarrow y_{l_i}\rightarrow1$. 0 The statement is not true in the sense that it is provable. The statement is true in the sense that as a definition, it is consistent (not contradictory, not paradoxical). Or more exactly, it is a special case of an generalized definition of summations. You are probably familiar with the definition $$\sum_{n=1}^\infty f(n) = \lim_{k\rightarrow \infty} ... 1 The question cited as 'possible duplicate' covers why \sum_{n=1}^{k}n may or may not be considered to be -1/12. However, there are some other important points to make:$$\sum_{n=1}^{k}n=\frac{k(k+1)}{2} \implies \frac{\infty(\infty+1)}{2}=\frac{-1}{12}$$This would only be true if k were a positive integer (otherwise, your sum, and more ... 0 Here what you can do. Solve as normal until you have the equation in the form ax^2+bx+c=0. This is a quadratic equation and using the quadratic formula you arrive at 13. Ignore the negative value as "n" cant be negative. 0 The relation between the number of terms and the first and last terms can be found by a_{f}=a_{o}+n*d Essentially, the last term would just be expressed as the the first term (your starting point) plus however many 'steps' (i.e. - each step would be 'd' or the common ratio) are needed to get to the last term. The number of steps, as always in an ... 0 To see that it's not a total order, take the counter-example provided by user Git Gud in the comments, i.e. take x_1=01 and x_2=10. Then there is no \beta, such that x_1.\beta = x_2. To prove that it is a partial order, you have to check reflexivity, antisymmetry and transitivity. You get reflexivity for free from the defnition. To check ... 1 The Fermat numbers 2^{2^n}+1 are pairwise relatively prime - this is not just an observation but is in fact provable. It's easy to turn this into a function f\colon \Bbb Z\to\Bbb N where the outputs are pairwise relatively prime:$$ f(n) = \begin{cases} 2^{2^{2n}}+1, &\text{if }n\ge0; \\ 2^{2^{2|n|-1}}+1, &\text{if }n\le-1. \end{cases} $$Or, if ... 0 Alternatively you can use the fact that the series \sum_{n=0}^\infty\frac{n}{4^n} converges by the ratio test. For n>0$$\frac{n+1}{4^{n+1}}\frac{4^n}{n}=\frac{1}{4}\bigg(1+\frac{1}n{}\bigg)\to \frac{1}{4}<1$$. Hence \frac{n}{4^n} \to 0 0 (1-x^2)f(x) \ge x means f(x) - \frac{x}{1-x^2} \ge 0 Now, f(x) - \frac{x}{1-x^2} = \frac{x}{1-x} + \frac{2x^2}{1-x^2} - \frac{x}{1-x^2} + \cdot\cdot\cdot \ge \frac{x+x^2 + 2x^2 - x}{1-x^2} = 3\frac{x^2}{1-x^2} \ge 0 I hope I didn't miss anything. 3 For the other way: Hint: If \sum\frac{x_n}{1+x_n} converges, then in particular x_n must converge to 0, and in particular there exists a constant C\geq0 such that x_n\leq C. Then$$\sum\frac{x_n}{1+x_n}\geq\sum\frac{x_n}{1+C}.$$3 As$$ 0 \le \frac{x_n}{1+x_n}\le x_n,\\ \sum {x_n}<\infty\Rightarrow \sum \frac{x_n}{1+x_n}<\infty $$Now suppose that \sum \frac{x_n}{1+x_n}<\infty . \frac{x_n}{1+x_n}\to 0, and so x_n\to 0, so there is a N such as n>N\Rightarrow x_n<\frac 12 and then$$ \frac{x_n}{1+x_n}>\frac 23 x_n $$so$$ \sum ... 1 Let$a_n=\frac{(n+3)!4^n}{5n^{n+2}}$. To use the ratio test, we want to calculate$\lim_{n\rightarrow\infty}a_{n+1}/a_n: \begin{align*} \lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}&=\lim_{n\rightarrow\infty}\frac{\frac{(n+4)!4^{n+1}}{5(n+1)^{n+3}}}{\frac{(n+3)!4^n}{5n^{n+2}}}\\ ... 0 Use the Stirling formula: $$\log\frac{{(n+3)!} (4^n)}{{5n}^{n+2}} = \log (n+3)! + n\log 4 - (n+2)\log n - \log 5 \\=\ (n+3)\log (n+3) - (n+3) + n\log 4 - (n+2)\log n+ O(\log n)\\ = n(\log 4-1) + O(\log n)\to\infty$$so the series is divergent. 1 Note that, thanks to the Stirling Formula: $$\log \frac{{(n+2)}^{n+1}}{(n+1)!} = (n+1) \log(n+2) - \log (n+1)! \sim n$$ So your series is divergent. 0 Now you can write this as\displaystyle \lim_{n\to\infty}\bigg(\frac{n+3}{n+2}\bigg)^{n+1}\cdot\frac{n+3}{n+2}=\lim_{n\to\infty}\frac{(1+\frac{3}{n})^n}{(1+\frac{2}{n})^n}\cdot\bigg(\frac{n+3}{n+2}\bigg)^2=\frac{e^3}{e^2}\cdot1=e,$or use$m=n+2$to get$\displaystyle ...

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$$a>1\Rightarrow\frac n{a^n}\to 0$$ Let us prove that there is a $C$ such as, for each $b>1$: $$\frac n{b^n}\leq C$$ if it is true for $n$, then $$\frac {n+1}{b^{n+1}} = \frac {n+1}{bn}\frac n{b^n} \leq C\frac {n+1}{bn}\leq C$$ as soon as $(b-1)n\ge 1$. Now take $$C = \max \{n{2^n}|(b-1)n\ge 1\}$$ We have $$\frac n{2^n}\leq C$$as soon ...

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Use induction to show that $n<2^n$. When $n=1$, the result is trivial. Now, assume that $2^n>n$ is true. Then, $$2^{n+1}=2\cdot2^n>2n=n+n>n+1\mbox{, as we are assuming n\geq1 (we've proved it for n=1).}$$ So, $$2^{n+1}>n+1$$ Thus, letting $m=n+1$ shows that the inequality is true for $m>1$. You need to show that ...

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Plugging in $1$ and not $2$ in the bernoulli inequality you get $$(1+1)^n> 1+n \implies 2^n > 1+n>n$$ and thus you have proven the hint, which indicates that $\frac{n}{2^n}<1$. Therefore $$\frac{n}{4^n}=\frac{n}{(2\cdot2)^n}=\frac{1}{2^n}\cdot \underbrace{\frac{n}{2^n}}_{< 1 }<\frac{1}{2^n}$$ where the RHS converges to $0$ as $n \to ... 1 To use the hint, use it on the numerator: $$\dfrac{n}{4^n} < \dfrac{2^n}{4^n} = \left(\dfrac{2}{4}\right)^n = \dfrac{1}{2^n} \to 0$$ You can also prove$n < 2^n$with induction. 0 @Marek, For me the hint was in http://oeis.org/A097321, from which the numerators for log(3) are 1,1,-2. Now log(2) having 1, -1 and log(3) having 1, 1, -2 suggests the pattern. 0 It does not follow. For example $$u_n=2^n+(-2)^n$$ has characteristic equation with roots$2$and$-2$; however$u_n=0$whenever$n$is odd, so$u_n$does not tend to$\infty$. 0 $$\frac \beta{k^\alpha} - \frac \beta{(k+1)^\alpha} \sim \frac {\alpha\beta}{k^{\alpha + 1}} = \frac 1{\sqrt{k}}$$if$\alpha = -1/2$and$\beta = -2$now as$\sum \frac 1{\sqrt{k}} = \infty$: $$\sum_{k=1}^N \frac 1{\sqrt{k}}\sim \sum_{k=1}^N \frac \beta{k^\alpha} - \frac \beta{(k+1)^\alpha} \sim -\frac \beta{N^\alpha} = 2\sqrt{N}$$ 1 Given$\varepsilon$, choose$n_0$such that for all$n\ge n_0$we have$|a_n-a|<\min\bigg(1 \frac{\varepsilon}{|a|^2+|a|(1+|a|)+(1+|a|)^2}\bigg)$so$|a_n| \le1+|a|\begin{align}|a_n^3-a^3|=|a-a_n||a^2+aa_n+a_n^2|\le |a_n-a|(|a|^2+|a||a_n|+|a_n|^2)\\ \le|a_n-a|\big[|a|^2+|a|(1+|a|)+(1+|a|)^2\big]\\<\varepsilon \end{align} 3 By looking at upper and lower Riemann sums on this decreasing function, you can bracket it by $$\int_1^{n} x^{-\frac 12}\ dx \lt \sum_{k=1}^nk^{-\frac 12} \lt \int_0^{n-1} x^{-\frac 12}\ dx\\2\sqrt n-2\lt \sum_{k=1}^nk^{-\frac 12} \lt2\sqrt {n-1}$$ 0 Since\alpha_n \to \alpha$there is an$n_0$such that$|\alpha_n - \alpha| < \sqrt[3]{\epsilon}$for every$n \geq n_0$. Now raise both sides to the$3$rd power, which is allowed as the function$x \mapsto x^3$is strictly increasing. 1 This is a point of definitions and naming conventions. My preferred definition is that when $$S_n = \sum_{i=0}^n a_i$$ we say that the sequence$S_n$is the series of term$a_n$. Both$S_n$and$a_n$are sequences. While $$S = \lim_{n\to \infty} S_n = \sum_{i=0}^\infty a_i$$ is the sum of the series of term$a_n$.$S$is a number (or$\pm \infty$). 1 I don't know what do you mean by "The sequence gets reset every$40$items", so I assume that$Q$is like $$P,P+2,P+4,P+6, \ldots ,P+2(M-1),P,P+2,P+4, \ldots$$ where$P=$starting even number and$M=$the number of items of$Q$between$2$starting numbers, so the greatest number in the sequence will be$P+2(M-1)$. Let$A$be the first element of the ... 5 Bump the exponent up again, so you get $$\sum_{k=1}^\infty \frac {(-1)^k}{k}\cdot\frac{x^{2k}}{4^k}.$$ Then at$x=-2$, the latter term vanishes like you want. 0 As per Daniel Fischer's hint, $$0 \leq s_n \leq \sum_{k = 1}^{m_n}\left[\left(t^{(n)}_k - t^{(n)}_{k - 1}\right)\underbrace{\max_{i \in \left\{1, \dots, m_n\right\}}\left(t^{(n)}_i - t^{(n)}_{i - 1}\right)}_{=: M_n}\right] = M_n\sum_{k = 1}^{m_n}\left(t^{(n)}_k - t^{(n)}_{k - 1}\right) = M_nt$$ Since,by assumption,$\lim_{n \rightarrow \infty} M_n = 0$, ... 0 HINT: As the numerator is cubic, we need to express$\displaystyle k^3+6k^2+11k+5$as$1\cdot(k+3)(k+2)(k+1)+b_1(k+3)(k+2)+b_2(k+3)+b_3$where$b_i$s are arbitrary constants Observe that we have to start with$k+3$as it is there in the factorial of the denominator. Had the numerator been square polynomial like$A k^2+Bk+Cwith the denominator ... 0 (The ideas appear also in the comments and other posts). Hint: \begin{align*}\frac{k^3+6k^2+11k+5}{(k+3)!}&=\frac{k^3+6k^2+11k+6-1}{(k+3)!}=\frac{(k+3)(k+2)(k+1)-1}{(k+3)(k+2)(k+1)k!}=\\ \\&=\frac{(k+3)(k+2)(k+1)}{(k+3)(k+2)(k+1)k!}-\frac{1}{(k+3)(k+2)(k+1)k!}=\\&=\frac{1}{k!}-\frac{1}{(k+3)!}\end{align*} Now, since ... 1 That series can be written: $$\sum\limits_{k=1}^\infty\left(\frac 1{k!}-\frac 1{(k+3)!}\right)$$ so it is(e-1)-(e-1-1/1-1/2-1/6)=5/3$1 Hint: For all$x\in \mathbb R$it holds that$e^x=\sum \limits_{n=0}^{\infty}\left(\dfrac {x^n}{n!}\right)$and, differentiating ,$e^x=\sum \limits_{n=1}^{\infty}\left(\dfrac {nx^{n-1}}{n!}\right)$. Differentiating again yields$e^x=\sum \limits_{n=2}^{\infty}\left(\dfrac{n(n-1)x^{n-2}}{n!}\right)$and again yields$e^x=\sum \limits_{n=3}^\infty ...

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Let P(i) denotes the i-th prime number, then when n is between P(i) and P(i+1)-1, $\pi$(n) doesn't change. Group these terms with the same sign together, we obtain a new series: $\sum \hat{a}(k)$. Now we have an alternating series. we use the Leibniz's theorem: "If the absolute value of $\hat{a}(k)$ decreases with k, and $\lim_{n\to\infty}\hat{a}(k)=0$ ...

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Suppose to the contrary that converges. Let $s_n$ denote the $n$-th partial sum. Since the serie converges, $(s_n)$ is a Cauchy sequence. Let $\varepsilon = 1/3$, then there is some $n_0$ such that $|s_q-s_p|< 1/3$ for all $q>p\ge n_0$. Let $q=2n_0$ and $p=n_0$. Then $$\frac{1}{3}>\bigg|\sum_{n=n_0+1}^{2n_0} ... 1 Alternatively you can show that (s_n) [the n-th partial sum] is a Cauchy sequence: Denote S_n the n-th partial sum for the absolute values of b_n, i.e., S_N =\sum_{n=0}^N |b_n|. Choose n_0 such that S_q-S_p<\varepsilon, for all q>p\ge n_0. Then$$|s_q-s_p|=\bigg|\sum_{n=p+1}^q b_n \bigg|\le\sum_{n=p+1}^q |b_n|=S_q-S_p< ...

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You know that if we have two convergent sequences (in the wide sense of the word) $\;\{a_n\}\;,\;\;\{b_n\}\;$ s.t. $\;a_n\le b_n\;$ , then $\;\lim a_n\le\lim b_n\;$ . Well, now apply this to the sequences of partial sums by means of the triangle inequality: $$\forall\,N\in\Bbb N\;,\;\;\left|\sum_{n=1}^na_n\right|\le\sum_{n=1}^N|a_n|\implies ... 0 Another (different) answer, by the Cauchy Condensation Test :$$\sum_{n=1}^\infty \frac{1}{n} < \infty \iff \sum_{n=1}^\infty 2^n \frac{1}{2^n} = \sum_{n=1}^\infty 1< \infty $$The latter is obviously divergent, therefore the former diverges. This is THE shortest proof there is. 0 I tried to solve this from scratch:$$ \\ T(n) = 2T(n - 1) + \frac{1}{n} \\ T(n) = 2(2T(n - 2) + \frac{1}{n - 1}) + \frac{1}{n} \\ T(n) = 2(2(2T(n - 3) + \frac{1}{n - 2}) + \frac{1}{n - 1}) + \frac{1}{n} = 2^3T(n - 3) +\frac{2^2}{n - 2}+\frac{2^1}{n - 1} + \frac{2^0}{n} \\ T(n) = 2^kT(n - k) + \sum_{i = 0}^{k - 1} \frac{2^i}{n - i} \\ \text{When } n - k = 0 ...

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There is an $r\ge a$ such that $$\frac{f(x)}{g(x)}\le 1 \quad \forall x \ge r.$$ It follows that $$0\le \int_a^\infty f(x)\,dx\le \int_a^rf(x)\,dx+\int_r^\infty g(x)\,dx<\infty.$$

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