Hot answers tagged

7

From $$ xS(x)=\sum_{n=0}^\infty\frac{x^{n+1}}{(n+1)!}, $$ just by a change of index, you get $$ xS(x)=\sum_{p=1}^\infty\frac{x^p}{p!}=\sum_{p=0}^\infty\frac{x^p}{p!}-1=e^x-1, $$ that is $$ S(x)=\frac{e^x-1}x, \quad x \neq0. $$


5

HINT: $$\sin(n)\sin\left(\frac{(-1)^n}{n^{1/4}}\right)=(-1)^n\sin(n)\sin\left(\frac{1}{n^{1/4}}\right)$$ Now, show that there exists a number $M$ (added hint: $M=\sec(1/2)$ suffices) such that for all $N$ $$\left|\sum_{n=1}^N (-1)^n\sin(n)\right|\le M$$ Finally, apply Dirichlet's Test noting that $\sin\left(\frac{1}{n^{1/4}}\right)>0$ and ...


4

The typical way to "skip terms" in a power series is to use roots of unity. Here, let $\zeta=e^{i\pi/3}$ be a primitive $6$th root of unity, so that $$ \frac16 \sum_{j=0}^5 (\zeta^\ell)^j = \begin{cases} 1, &\text{if } 6\mid \ell, \\ 0, &\text{otherwise.} \end{cases} $$ Therefore, setting $\alpha=e^{i\pi/6}$, for any $0\le k\le 5$, \begin{align*} ...


4

May be, we could use generalized harmonic numbers since $$\sum_{n=1}^x {1\over \sqrt{n}}=H_x^{\left(\frac{1}{2}\right)}$$ and use the asymptotics for large values of $n$ $$H_x^{\left(\frac{1}{2}\right)}=2 \sqrt{x}+\zeta \left(\frac{1}{2}\right)+\frac 1 {2\sqrt x}+O\left(\frac{1}{x^{3/2}}\right)$$


4

Yes, there is. As it is a series with positive terms, you can use equivalents: $$\log(2n+1)-\log(2n)=\log\Bigl(1+\frac1{2n}\Bigr)\sim_\infty\frac1{2n},$$ which diverges.


3

use that $$\sum_{k=1}^n k^{-1/2} = \int_{1-\epsilon}^{n+\epsilon} \left(\sum_{k=1}^n \delta(x-k)\right) x^{-1/2} dx = n^{1/2} + \frac{1}{2}\int_1^n \lfloor x \rfloor x^{-3/2} dx$$ (some sort of integration by parts, see Abel's summation formula ) while $$ n^{1/2} = 1+\frac{1}{2}\int_1^n x^{-1/2} dx$$ hence $$A_n = 2 n^{1/2}-\sum_{k=1}^n k^{-1/2} = 1 ...


3

Too lazy to whip up anything fancy, but program z2 use ISO_FORTRAN_ENV,only:wp=>REAL128 real(wp) zeta,x integer k, M M = 100000000 zeta = sum([(1/sqrt(real(k,wp)),k=1,M)])- & 2*sqrt(M+0.5_wp)-1/(48*sqrt(M+0.5_wp)**3) write(*,*) zeta end program z2 Spits out $-1.46035450880958681288949915247298$, which has $29$ digits of ...


3

$$\forall x\in\mathbb{C}:\|x\|<1,\qquad\sum_{n\geq 1}x^n = \frac{x}{1-x}\tag{1} $$ Now we apply twice the operator $xD: f(x)\mapsto x\cdot f'(x) $ to get: $$\forall x\in\mathbb{C}:\|x\|<1,\qquad\sum_{n\geq 1}n^2\,x^n = \frac{x(1+x)}{(1-x)^3}\tag{2} $$ and by evaluating at $x=\frac{2}{5}$: $$\sum_{n\geq 1}n^2\left(\frac{2}{5}\right)^n = ...


3

Using $\;\sin\frac1n\le\frac1n\;$ , you get: $$\left|\frac{\cos n\sin\frac1n}{n}\right|\le\frac1{n^2}$$ annd the comparison test gives you absolute convoergence ( of course, we also used $\;|\cos n|\le1\;$).


3

The answer is negative. In fact if we define the Baire class $B_\alpha$ for countable ordinals $\alpha$ by saying $B_0=C(I)$, $B_{\alpha+1}$ is the set of pointwise limits of sequences in $B_\alpha$, and $B_\alpha=\bigcup_{\beta<\alpha}B_\beta$ for limit ordinals $\alpha$ then all the $B_\alpha$ are distinct. Or so I've read; don't ask me to prove it. ...


3

Fine as they are, the other answers gloss over an important nuance: In a comparison of (positive) sequences, if $a_n\le b_n$ and $\lim_{n\to\infty}b_n=L$, we cannot conclude that $\lim_{n\to\infty}a_n=L$. In fact, we cannot even conclude that $\lim_{n\to\infty}a_n$ exists (unless $L=0$). We need to squeeze $a_n$ between two sequences that both converge to ...


3

First, note that there exists $\delta$ such that $|\sin^4(x)|<\delta$ implies $|\sin^4(x+1)|>\delta$. To see this, draw a circle. The angles $x$ for which $|\sin(x)|<\delta$ correspond so arcs around $(1,0)$ (at angle $0$) and $(-1,0)$ (at angle $\pi$), so if these arcs are small enough, which corresponds to $\delta$ being small, then adding $1$ to ...


3

Note that $$ \sin^4 n = (\sin^2 n)^2 = \left( \frac{1 - \cos 2n }2 \right)^2 = \frac{1 - 2\cos 2n + \cos^2 2n}4 $$ $$ = \frac{1-2\cos 2n}4 + \frac{1+2\cos 4n}8 = \frac38-\frac{\cos 2n}2+\frac{\cos 4n}4. $$ We split the sum into three parts: $$ \sum_{n=1}^{\infty} \frac{\sin^4 n }{\ln(n+\frac1n)} = \sum_{n=1}^{\infty} \frac38 \frac1{\ln(n+\frac1n)} - ...


3

Without too many details: $|(1+i/n)^n| = (1 + \frac{1}{n^2})^{n/2}$ $\arg(1 + i/n)^n = n\tan^{-1} 1/n$ as $n \to \infty$, the modulus goes to $1$ and the argument goes to $1$. This is exactly $e^i$.


3

Hint: By mean value theorem, for each $n\in\mathbb{N}$ there exists $\xi_n\in(2n,2n+1)$ such that $$\log(2n+1)-\log(2n)=\frac{1}{\xi_n}\geq\frac{1}{2n+1}\geq\frac{1}{3n}$$


3

The series converges: Using $\sin x = x - \frac{x^3}{6} + o(x^4)$ around $0$, we get that when $n\to \infty$ $$ \sin\left(\frac{\sin(3n)}{\sqrt{n}} + \frac{1}{n\ln^2(n)}\right) = \underbrace{\frac{\sin(3n)}{\sqrt{n}}}_{a_n} + \underbrace{\frac{1}{n\ln^2(n)} + O\left(\frac{1}{n^{3/2}}\right)}_{b_n}. $$ The series $\sum_{n=2}^\infty (-1)^n b_n$ converges ...


3

The integral of the question $1$ is not difficult to evaluate. We have $$\int_{0}^{1}\left(\left(y+\frac{1}{y}\right)\log\left(1+y^{2}\right)-y\right)dy=\int_{0}^{1}y\log\left(1+y^{2}\right)dy+\int_{0}^{1}\frac{\log\left(1+y^{2}\right)}{y}dy-\int_{0}^{1}ydy. $$ For the first integral take $y^{2}+1=u $ to get ...


2

Multiply the given relation by a factor of $\frac{1}{2}$ and add it to the original sequence. Now we get: $$L = 1 - \frac{1}{2} + \frac 13 - \frac 14 + \frac 15 - \frac 16 + \frac 17 - \frac 18 + ...$$ $$\frac 12 L = 0 + \frac 12 + 0 - \frac 14 + 0 + \frac 16 + 0 - \frac 18+ ...$$ $$\frac 32 L = 1 + 0 + \frac 13 - \frac 12 + \frac 15 + 0 + \frac 17 - ...


2

Observe that $$a_n:=\frac{ne^{-n^2}}{e^{-n}+4}=\frac{ne^n}{4e^{n^2+n}+e^{n^2}}\implies$$ $$\sqrt[n]{a_n}=\frac{\sqrt[n]n\,e}{\sqrt[n]{4e^{n^2+n}+e^{n^2}}}\le\frac{\sqrt[n]n}{e^n}e\xrightarrow[n\to\infty]{}0$$ so the series converges.


2

Hint. One has, for $n \geq1$, $$ 0\leq\frac{ne^{-n^2}}{e^{-n}+4}\leq ne^{-n^2}. $$


2

Yes, the important thing is that it is decreasing after a while at least. This is because a finite number of terms isn't relevant for the convergence of the serie.


2

Let $(x_n)$ be a convergent sequence in $M$ with limit $a$. Then we can consider the sequence $x_1, a, x_2, a, x_3, a, \ldots$ which intertwines $(x_n)$ with the constant sequence $(a)$. This sequence converges to $a$ as well, so by assumption, the sequence $f(x_1), f(a), f(x_2), f(a), \ldots$ converges in $N$. Since every other term is $f(a)$, the only ...


2

Note that $\cos kz = \mathrm{Re}(\cos kz + i \sin kz) = \mathrm{Re}(e^{ikz})$. Thus: $$ \sum_{k \geq 1} e^{-tk} \cos kz = \sum_{k \geq 1} \mathrm{Re}(e^{-tk})\mathrm{Re}(e^{ikz}) = \sum_{k \geq 1} \mathrm{Re}(e^{k(-t + iz)}) = \mathrm{Re}\left(\sum_{k \geq 1} \left(e^{-t + iz}\right)^k\right) $$ which is a geometric series.


2

By the triangle inequality, $$\int_{B_n} |f_n-f| \mathop{d\mu} \le \int_{B_n} (|f_n|+|f|) \mathop{d\mu} \le 2M \int_{B_n} \mathop{d\mu}$$ where the last inequality is due to $|f_n| \le M$ and $|f| \le M$.


2

for $x=\pi$, the series is $$ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \sin(n\pi) = \sum_{n=1}^{\infty} 0 = 0 $$ so the series converges to 0 at $\pi$. We expect this because 0 is the midpoint between $g(-\pi)$ and $g(\pi)$


2

The two series $$ f(z)=\sum \frac{z^n}{n},\quad g(z)=\sum (-1)^n\frac{z^n}{n} $$ are a counterexample to your conjecture, because $f(z)$ converges everywhere on the unit disk except at $z=1$ while $g(z)$ converges everywhere on the unit disk except at $z=-1$ (note, indeed, that $f(-z)=g(z)$). The answer to your second question is affirmative, as one can ...


2

One may consider, $\dfrac1{x_n^2}=\pi n$, with $n=1,2,3,\ldots$, giving $$ \lim _{x_n\to 0}\:\frac{\sin\left(\frac{1}{x_n^2}\right)}{x_n^2}=\lim _{n\to +\infty}\pi n \times \sin(\pi n)=0 $$ and one may consider, $\dfrac1{y_n^2}=(4n+1)\dfrac{\pi}2 $, with $n=0,1,2,\ldots$, giving $$ \lim _{y_n\to ...


2

break it up. $a_n=\sum \limits_{k=1}^{n} \frac{2^kn}{2^{k+1}n^2+2^kk} + \sum\limits_{k=1}^{n} \frac{2n^2}{2^{k+1}n^2+2^kk}+\sum\limits_{k=1}^{n} \frac{k}{2^{k+1}n^2+2^kk}.$ Since every element of each of those series is greater than zero, the series for $a_n$ converges iff all three of the above converge, and diverges if any of the above diverge. $\sum ...


2

Let $f_n(x) = e^{-x^2} n \sin\left(\dfrac{x}{n}\right)\cdot \chi_{[0,n^2]}(x)$. On the interval $[0,1]$, this is a monotone increasing sequence of functions, and MCT can be applied. On the interval $[1,\infty)$, you have $$ \int^{n^2}_1 e^{-x^2} n \sin\left(\frac{x}{n}\right) \, dx = \int\limits_{[1,\infty)} f_n(x)\,dx $$ and $$ |f_n(x)| \le xe^{-x^2} ...


2

For $\;|x|<1\;$ : $$\frac1{1-x^2}=\sum_{n=0}^\infty x^{2n}\stackrel{\text{diff.}}\implies\frac{2x}{(1-x^2)^2}=\sum_{n=1}^\infty 2n\,x^{2n-1}$$ Now just do a little cosmetics to the above and get your answer.



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