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17

We may obtain a closed form for this integral. Proposition. Let $k=1,2,3,\ldots$. Then $$\int_0^1 \frac{B_{2k+1}\left(\left\{ 1/x \right\}\right)}{x}dx=(-1)^{k+1}\frac{(2k+1)!}{(2\pi)^{2k+1}}\pi \: \zeta(2k+1)-\sum_{j=0}^{2k}\!\frac{ {{2k+1}\choose j} B_j}{2k+1-j} \quad (*)$$ Proof. Recall the celebrated Fourier expansion (see this very nice paper, ...

8

Hint $2^n+3^n\leq 2\cdot 3^n$, hence your sum is bounded by $$\sum_{n} \frac{2\cdot 3^n}{4^n}...$$

7

Let $f(n)=[\prod_{i=0}^n (1+r/n)]^{1/n}$ Then, $\ln f(n)=1/n\sum_{r=0}^n\ln (1+r/n)\Rightarrow \lim_{n\to \infty}\ln f(n)=\int_{0}^1 \ln (x+1) dx=\ln 2-\int_{0}^1 \dfrac{x}{x+1}dx=2\ln 2-1\Rightarrow \lim_{n\to \infty} f(n) =4/e$

5

Let $s_n=a_1+\cdots+a_n\to a$. Then $$\sum_{k=1}^n ka_k=\sum_{k=1}^n k(s_k-s_{k-1})=\sum_{k=1}^n ks_k-\sum_{k=1}^{n-1}(k+1)s_k=ns_n-\sum_{k=1}^{n-1}s_k.$$ Hence $$\frac{1}{n}\sum_{k=1}^n ka_k=s_n-\frac{n-1}{n}\cdot\frac{1}{n-1}\sum_{k=1}^{n-1}s_k\to a-a=0.$$ We have used the fact that: If $b_n\to b$, then so does $\,\,\dfrac{b_1+\cdots+b_n}{n}$. Note. ...

4

If $f_n,f\geq 0$ then you can consider the sequence $$g_n=\min(f_n,f)=\frac{1}{2}(f_n+f-|f_n-f|)$$ which is bounded by $f$ and and converges pointwise a.e. to $f$. Then use the dominated convergence to obtain the claim. Generally, if you allow $f_n,f$ to attain values in $\mathbb R$, the claim is not true. Consider the functions $f_n= n ... 3 Hint: $$\ln \left(\dfrac{k+2}{k+1}\right)=\ln (k+2)-\ln(k+1)$$ Now if you expand the sum (as you already have) you will see that this is a telescoping series. 3 Hints: Using l'Hospital, for example, show that$\;\log x\le x^\epsilon\;\;\;\forall\,\epsilon>0\;$, for$\;x>0\;$big enough And now use direct comparison test $$\frac{\log n}{n^2}\le\frac{n^{1/2}}{n^2}=\frac1{n^{3/2}}$$ 3 The series$\displaystyle \sum_{n \geq 1} \frac{1}{n^{\alpha}}$converges if and only if$\alpha > 1$. Note that : $$\frac{\ln(n)}{n^{2}} n^{\alpha} = n^{\alpha-2} \ln(n).$$ For any$\alpha$such that$\alpha > 1$and$\alpha-2 < 0$($\displaystyle \alpha = \frac{3}{2}$works!), we have : $$\lim \limits_{n \to +\infty} n^{\alpha-2}\ln(n) = 0 ... 3 Part 1. The series \sum_{n=0}^{\infty}t^n is convergent if and only if |t|<1 giving$$\sum_{n=0}^{\infty}t^n =\frac{1}{1-t}. \tag1$$Thus putting t:=\frac{x+3}{2} in (1), we see that we must have |x+3|<2 equivalently -5<x<-1, giving$$\sum_{n=0}^{\infty}\left(\frac{x+3}{2}\right)^n =\frac{1}{1-\frac{x+3}{2}}=-\frac{2}{x+1}. ... 3 Notice that $$\sqrt[p]{\sum_{k=3}^{\infty} (k - H_k)^{-p}}$$ is simply the$\ell^p$-norm of the sequence $$\frac{1}{3-H_3}, \frac{1}{4-H_4}, \ldots, \frac{1}{k-H_k}, \ldots$$ Taking the limit as$p\to \infty$, assuming it exists, gives us the$\ell^\infty$-norm of this sequence. (See The$ l^{\infty} $-norm is equal to the limit of the$ l^{p} $-norms. for ... 3 This looks great. You checked that$f(t)$was monotone decreasing, then showed that the improper integral diverged. However, you could've gotten to your result a lot faster with the comparison test. For all$n > e$, we have: $$\frac{\ln(n)}{n} > \frac{1}{n}$$ Since$\displaystyle \sum_{n=1}^\infty \frac{1}{n}$diverges, then... 3 Note that $$\frac{1}{k}<\frac{1}{k-e^{-k}},$$ and as$\sum_{k=1}^\infty\frac{1}{k}$diverges, so does$\sum_{k=1}^\infty\frac{1}{k-e^{-k}}$. 3 We shall use that: Fact.$\displaystyle\lim_{n\to\infty}\left(\frac{n^n}{n!}\right)^{\!1/n} \!\!\!\!\!\longrightarrow\mathrm{e}.$Then $$\left(\frac{(2n)!}{n^n n!}\right)^{1/n}=\left(\frac{2^{2n}\frac{n^n}{n!}}{\frac{(2n)^{2n}}{(2n)!}}\right)^{\!1/n}=4\frac{\left(\frac{n^n}{n!}\right)^{1/n}}{\left(\frac{(2n)^{2n}}{(2n)!}\right)^{1/n}} ... 2 Hint: b_k \geq \frac{a_1}{k}, because a_n > 0 for all n. 2 On one hand, the series is strictly increasing. On the other hand, n^3+2\le2n^3 and n^4+3n^2+1 \ge n^4, implying t_n\le\ldots~\Big(I'll let you fill in the dots, and draw the conclusion for yourself\Big). 2 Let a_{n} = \frac{\sqrt{n^{3}+2}}{n^{4}+3n^{2} + 1} and let b_{n} = \frac{1}{n^{\alpha}} for some positive \alpha. Consider$$\lim_{n\rightarrow\infty} \frac{a_{n}}{b_{n}} = \lim_{n\rightarrow \infty} \frac{\frac{\sqrt{n^{3}+2}}{n^{4}+3n^{2} + 1}}{\frac{1}{n^{\alpha}}} = \lim_{n\rightarrow \infty} \frac{n^{\alpha}\sqrt{n^{3}+2}}{n^{4}+3n^{2} + 1} = ... 2 Hint: For$x$large enough,$p(x)<x^{n+1}$. What can we say about$\frac{e^x}{x^{n+1}}$as$x\to\infty$? 2 No. $$\frac{e^x}{p(x)}=1+\frac{\sum\limits_{k=n+1}^\infty \frac {x^k}{k!}}{p(x)}\stackrel{x\to\infty}\to\infty$$ 2 For any polynomial$P$, the limit $$\lim_{x\to\infty}\frac{e^x}{P(x)}$$ is equal to plus or minus$\infty$(depending on the sign of the leading coefficient of$P$). This statement does not change just because you took the polynomial approximation for$e^x$as your polynomial$P$. 2 Hint: Consider$x_n = \frac{1}{n(\log n)^r}$for$r>0$. Then $$x_n \log x_n^{-1} = \frac{1}{n(\log n)^r} \log(n\log(n)^r)) > \frac{1}{n(\log n)^{r-1}}$$ The integral test can be used to find the range of$r$for which the series$\sum \frac{1}{n(\log n)^r}$converges. 2 The sequence is derived from an application of Newton's method for$f(x) = x^3 - 2$. So clearly it'd converge to the real root$\sqrt[3]{2}$, given you choose an appropriate value for$x_0$: $$x_{n+1} = x_n -\frac{f(x_n)}{f'(x_n)}=x_n-\frac{x_n^3-2}{3x_n^2 }=\frac{2-x_n^3+3x_n^3}{3x_n^2 }=\frac{2x_n^3+2}{3x_n^2}$$ For the second part you should find a ... 2 Hint: the numerator is bounded from above and below, but denominator contains$n^4$, which is the dominant term as$n\to+\infty$. Hence your series is comparable with $$\sum_n \frac{1}{n^4} ...$$ Let$a_n$denote the$n$-th term in the series. The absolute value of you numerator does not exceed $$a(10000\varepsilon \delta_1+10000\varepsilon \delta_2+4102 ... 2 I should start with 2) using induction. 2) If 1<a_n\leq 2 then:$$a_{n+1}=\frac{a_n+1}{2}>\frac{1+1}{2}=1$$and$$a_{n+1}=\frac{a_n+1}{2}\leq\frac{2+1}{2}\leq 2$$1) From a_n>1 it follows directly that:$$a_{n+1}=\frac{a_n+1}{2}<a_n$$A bounded monotone sequence is convergent and we can find its limit a on base of the relation ... 2 Using this method which's called Cauchy condensation we get$$\sum_{k\ge1}\frac{2^k}{2^k\ln 2^k\ln\ln(2^k)}=\frac1{\ln2}\sum_{k\ge1}\frac1{k\ln(k\ln2)}\sim\frac1{\ln2}\sum_{k\ge1}\frac1{k\ln(k)}$$so the series$$\sum_{n\ge1}\frac1{n\ln n\ln(n\ln n)}$$is divergent. Can you now solve the second series? 2 Observe that: \displaystyle \lim_{n \to \infty} \dfrac{\ln\left(1+\dfrac{1}{n+1}\right)}{\dfrac{1}{n+1}} = 1, and the harmonic series \displaystyle \sum_{n=1}^\infty \dfrac{1}{n+1} diverges, so the series \displaystyle \sum_{n=1}^\infty \ln\left(1+\dfrac{1}{n+1}\right) also diverges. 2$$P(|X_n-X|\geqslant\varepsilon)\leqslant\varepsilon^{-1}E(|X_n-X|)$$2 Here is a solution using Cauchy's inequality and Stolz–Cesàro theorem. By Stolz–Cesàro theorem, we need to prove$$\lim\dfrac{1}{n^2}\sum_{k=1}^n a_k^2 = \lim \dfrac{a^2_{n+1}}{2n+1} =0$$provided the second limit exists. Let$b_k = a_{k+1}- a_k$with$b_0 = a_1$, we have$\sum_k b_k^2 < \infty$and$a_{n+1} = \sum_{k=0}^{n}b_k. \begin{align} ... 2 What you write is the probability ofP(\sum_{k=1}^nX_n \geq \lfloor \frac{n}{2}\rfloor)$, where$X_i$'s are i.i.d Bernoulli variable with parameter$Q$.$P(\sum_{k=1}^nX_n \geq \lfloor \frac{n}{2}\rfloor) = P(\dfrac{\sum_{k=1}^nX_n -nQ}{\sqrt{n}} \geq \dfrac{\lfloor \frac{n}{2}\rfloor -nQ}{\sqrt{n}})and \begin{align} \lim_{n\to \infty} \dfrac{\lfloor ... 2 Your proof is OK. It means for large enoughn$,$\frac{ln(n)}{n^2}\leq \frac{1}{2}\frac{1}{n^\frac{3}{2}}$, where the LHS is convergent series. You can also use Cauchy Condensation Test. Then series converges iff$\sum_{n=1}^{∞} 2^n\frac{ln(2^n)}{(2^n)^2}=\sum_{n=1}^{∞} n\frac{ln2}{(2^n)}$, which is convergent. 2 Based on your statement of the problem, you don't assume that$X$is closed. So I don't think it is clear on the front end that the limit of a sequence is an element of$X$. Instead here is a hint. Assume$(y_k)_{k=1}^\infty$is a cauchy sequence. Express each$y_k = a_{1,k}e_1 + \cdots a_{n,k}e_n\$ using your basis. Use the definition of cauchy to prove ...

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