New answers tagged

0

Consider the function $$\frac{x^{10}}{x^2+1} \equiv 1-x^2+x^4-x^6+x^8-\frac{1}{x^2+1}$$ The integral: $$J = \int_0^1 \frac{x^{10}}{x^2+1} \text{d}x = \int_0^1 1-x^2+x^4-x^6+x^8-\frac{1}{x^2+1} \text{d}x$$ is clearly positive. Since $1\leqslant x^2+1 \leqslant 2$, we have: $$\frac{1}{2} \int_0^1 x^{10} \text{d}x < J < \int_0^1 x^{10} \text{d}x$$ ...


1

Simple answers above. The alternative is as follows: We know that $$1-\frac13+\frac15-\frac17+\frac19=\frac{263}{315}$$ which can be calculated by hand. We also know that $$\frac\pi4<\frac{3.2}4=0.8\quad\text{whereas}\quad\frac{263}{315}>\frac{252}{315}=0.8$$ so the inequality is equivalent to $$\left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \...


1

Compare the two sequences: $$\color{red}1,2,3,\color{red}{2^2},5,6,7,8,\color{red}{3^2},10,...,\underbrace{\color{red}{44^2}}_{1936},1937,1938,...,2010,\color{blue}{2011},2012,...,\underbrace{\color{green}{45^2}}_{2025},...,\underbrace{\color{green}{46^2}}_{2116},...\\ 2,3,5,6,7,8,10,...,2011,\underbrace{2012}_{\color{red}1},\underbrace{2013}_{\color{red}{2^...


1

$a_n = \frac{1}{n-1}$ where $n\ge 2$. It converges to $0$ as $n\to \infty$. It is bounded by $1$. Note that the sequence is not defined for $n=1$ as you will encounter with the zero division problem.


1

In the following, we derive Abel's test and Dirichlet's test for testing the convergence of infinite series in the form $\sum_{n=1}^{\infty}a_{n}b_{n}$. Firstly, we introduce Abel's transform, which is the discrete version of integration-by-part. Then we apply it to the difference of partial sums $S_{n+k}-S_{n}$. Section 1: Abel's transform. Let $(a_{n})$ ...


2

you can actually get a stronger result such as $$\left\vert\frac{\pi}{4} - \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9}\right)\right\vert < 1/11 $$ because the error in a convergent alternating series is less than the absolute value of the first missing term which in this case is $1/11.$ The alternating series is simply the Taylor ...


12

Use the series expansion of arctangent and the fact that for alternating series the rest is less than the first omitted term.


3

Your first error is swapping the indexes $n$ and $k$ at the start of the second line, since the $n=1$ term is $\left(\sum_{k=1}^\infty\frac{1}{1^{2k}}\right)-1=\left(\sum_{k=1}^\infty1\right)-1$, which is undefined. This is compounded by the lack of brackets, which makes it unclear what is inside the sum. Your second error is also in the second line, where ...


0

Take a look at the following picture : and see the connection with the issue in the so-called Argand plane (knowing that what is represented is a partial sum of the series). It remains to prove that this "inward spiraling" movement terminates as a limit point... therefore, I don't say that this graphics constitue a "proof without words" ; it just ...


2

Using $~\displaystyle\prod\limits_{k=1}^\infty\left(1-\left(\frac{x}{k}\right)^2\right)=\frac{\sin(\pi x)}{\pi x}~$ we have: $$\sum\limits_{k=1}^\infty\frac{x^{2k}}{2k}\zeta(2k) = -\frac{1}{2}\ln\frac{\sin(\pi x)}{\pi x}~,~~|x|<1$$ Differentiation and then substraction of $~\displaystyle\sum\limits_{k=1}^\infty x^{2k-1} = \frac{x}{1-x^2}~$ and after ...


4

I am going to use the following results : \begin{align} V_1=\sum_{n=1}^\infty\frac{(-1)^nH_n^{(4)}}{n}&=\frac78\ln2\zeta(4)+\frac38\zeta(2)\zeta(3)-2\zeta(5)\\ V_2=\int_0^{1/2}\frac{\ln^3(1-x)}{x(1-x)}\ dx&=6\operatorname{Li}_4\left(\frac12\right)-6\zeta(4)+\frac{21}{4}\ln2\zeta(3)-\frac32\ln^22\zeta(2)+\frac14\ln^42\\ V_3=\int_0^{1/2}\frac{\ln^3(1-...


2

Hint Dirichlet’s test may be your friend.


4

It is known that $\lim\limits_{y \to 0}\frac{\sin y}{y} =1$. Take $\epsilon >0$. You can pick up $\delta >0$ such that $\left\vert \frac{\sin y}{y} -1 \right\vert \le \epsilon$ for $\vert y \vert \le \delta$. For $n >a/\delta$ and $\vert x \vert \le a$, you have $\left\vert \frac{x}{n} \right\vert \le \delta$ and therefore $$\left\vert x \frac{\...


2

Note that, assuming that $x\in[-a,a]$,$$\left\lvert n\sin\left(\frac xn\right)-x\right\rvert=\left\lvert x\frac{\sin\left(\frac xn\right)}{\frac xn}-x\right\rvert=\lvert x\rvert\cdot\left\lvert\frac{\sin\left(\frac xn\right)}{\frac xn}-1\right\rvert\leqslant a\left\lvert\frac{\sin\left(\frac xn\right)}{\frac xn}-1\right\rvert.$$Now, use the fact that $\lim_{...


1

Just scale them. $A < B < C$ so $B$ is some proportion of the distance between $A$ and $C$. Just choose each $b_n$ so it is the same proportion between $a_n$ and $c_n$. ======== details ==== In other words $(B-A) = k*(C-A)$ for some $k; 0< k < 1$, so $k = \frac {B-A}{C-A}$ and $B = A + k*(C-A)= (1-k)A + kC$. Let $b_n = (1-k)a_n + k{C-A}c_n= ...


3

You want $b_n=\alpha a_n+(1-\alpha)c_n$, where $\alpha\in(0,1)$, with $B=\alpha A+(1-\alpha)C$. Solving this last equation for $\alpha$ yields $\alpha=(C-B)/(C-A)$.


1

If for all $n \in \mathbb N$ $0\le a_n < c_n$, $$b_n =(1-\lambda)a_n + \lambda c_n$$ will work where $\lambda \in (0,1)$ is such that $B=(1-\lambda)A+ \lambda C$.


1

Hint: Note that $$ \sum_{n=1}^\infty \frac{z^n}{1 + z^{2n}} \leq \sum_{n=1}^\infty \frac{1}{z^n} $$


7

Actually, for every $n>1$, $$\sum_{m=2}^{\infty}n^{-m}=\frac{1}{n(n-1)},$$ and $$\sum_{n=2}^{\infty}\frac{1}{n(n-1)}=1$$


1

Hint: $d(x_{n_k}, x_{n_p}) \leq \frac 1 {2^{k}}+\frac 1 {2^{k+1}}+...+\frac 1 {2^{p-1}}$ for $k <p$. The converse is immediate from definition of a Cauchy sequence.


2

Using Negative binomial theorem: $$\frac{1+t}{(1-t)^3}=(1+t)(1-t)^{-3}=(1+t)\sum_{k=0}^{\infty}{-3\choose k}(-t)^k=\\ \color{blue}{(1+t)\sum_{k=0}^{\infty}{2+k\choose k}t^k}=\sum_{k=0}^{\infty}\left[\color{red}{{2+k\choose k}+{1+k\choose k-1}}\right]t^k= \sum_{k=0}^\infty (1+k)^2t^k.$$ Note:


-1

Considering that $$ f(x) = \sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{\cdots}}}} $$ or $$ f(x) = \sqrt{1+x f(x+1)} $$ or $$ f^2(x) = 1+xf(x+1) $$ this gives us the possibility of calculating solutions with the structure $f(x) = a x + b$ or equating $$ (ax+b)^2 = 1+x(a(x+1)+b) $$ or $$ (1-b^2)+(a+b-2ab)x+a(a-1)x^2 = 0\;\;\forall x \Rightarrow a = b= ...


0

Its a hint Split the series to two. One is sum of cubic of odd. And other one is 3*(sum of square of even )


2

One can also express the area (as noted in the question) as a sum of trapezoids, obtained by dropping perpendicular lines from $E$, $G$, $I$, $K$, ... to the $x$-axis. The vertical bases of $k$-th trapezoid measure $r^k$ and $r^{k+1}$, while its height is $(1-r)\sqrt{1-r^{2k}}$. Hence total area $S$ can also be computed from: $$ S=\sum_{k=1}^\infty{1\over2}(...


1

Details for your comment above: $n$ is even: $$\sum_{1}^{n/2} (2k-1)^{3} +3 \sum_{1}^{n/2} (2k)^{2}$$ Example: $$1^3 + 3\cdot 2^2 = \sum_{k=1}^{2/2}(2k-1)^3+3\sum_{k=1}^{2/2}(2k)^2$$ $n$ is odd: $$\sum_{1}^{(n+1)/2} (2k-1)^{3} +3 \sum_{1}^{(n+1)/2-1} (2k)^{2}$$ Example: $$1^3 + 3\cdot 2^2 + 3^3 = \sum_{k=1}^{(3+1)/2}(2k-1)^3+3\sum_{k=1}^{(3+1)/2-1}(2k)^2$$


0

Hint $$\sum_{n=0}^{\infty} (2n+1)x^n=2x\sum_{n=0}^{\infty} nx^{n-1}+\sum_{n=0}^{\infty} x^{n}=2x \left(\sum_{n=0}^{\infty} x^{n} \right)'+\left(\sum_{n=0}^{\infty} x^{n} \right)$$


2

HINT When $n = 2m$ is even, both sums have the same amount of terms, $n/2 = m$ each. When $n = 2m-1$ is odd, the left sum has one more term than the right, so there must be $m$ terms in the left and $m-1$ in the right. Also notice that the even $n$ sum and the odd $n$ sum are different by just one last term in the right sum.


0

Hint Start writing $$(n+1)^2 x^n=(n^2+2n+1)x^n=n(n-1)x^n+3n x^n+x^n$$So, $$\sum_{n=0}^{\infty} (n+1)^2 x^n=x^2\sum_{n=0}^{\infty}n(n-1)x^{n-2}+3x\sum_{n=0}^{\infty}nx^{n-1}+\sum_{n=0}^{\infty}x^{n}$$ that is to say $$x^2\left(\sum_{n=0}^{\infty}x^{n} \right)''+3x\left(\sum_{n=0}^{\infty}x^{n} \right)'+\left(\sum_{n=0}^{\infty}x^{n} \right)$$


4

Let $S_n = \epsilon_1 + \ldots + \epsilon_n$, and apply summation by parts to get$$\sum_{j=1}^n \epsilon_jp_j = S_n p_n + \sum_{j=1}^{n-1} S_j(p_j - p_{j+1}).$$ If $\liminf S_n/n = \beta > 0$, then there exists $N$ such that $S_n/n > \beta/2$ for all $n > N$. Using the hypothesis that $p_n$ decreases and is positive since it converges to $0$ we ...


0

Let $f_m(x) =\sum_{n=0}^{\infty} n^mx^n $, so that $f_0(x) =\dfrac1{1-x} $, and $f_1(x) =\dfrac{x}{(1-x)^2} $. I will show that $$f_m(x) =\dfrac{(-1)^m-\sum_{k=0}^{m-1}\binom{m}{k}(-1)^{m-k}f_k(x)}{1-x} $$ and $$f_m(x) =\dfrac{x}{1-x}\sum_{k=0}^{m-1}\binom{m}{k}f_k(x) $$ First, $f_m(x)-xf_m(x) =(1-x)f_m(x) $ and $\begin{array}\\ f_m(x)-xf_m(x) &=\sum_{...


1

Without calculus. Start with $f_0(x) =\sum_{n=0}^{\infty} x^n =\dfrac1{1-x} $. If $f_1(x) =\sum_{n=0}^{\infty} nx^n $, then $f_1(x)-xf_1(x) =(1-x)f_1(x) $ and $\begin{array}\\ f_1(x)-xf_1(x) &=\sum_{n=0}^{\infty} nx^n-x\sum_{n=0}^{\infty} nx^n\\ &=\sum_{n=0}^{\infty} nx^n-\sum_{n=0}^{\infty} nx^{n+1}\\ &=\sum_{n=0}^{\infty} nx^n-\sum_{n=1}^{\...


0

$$\sum_{n=0}^{\infty} (n+1)^2 x^n = \sum\limits_{u=0}^{\infty}u^2x^{u-1} = f(x)$$ $\sum\limits_{u=0}^{\infty}ux^{u} = \int f(x)$ $x\sum\limits_{u=0}^{\infty}ux^{u-1} = \int f(x)$ $\sum\limits_{u=0}^{\infty}x^{u} = \int \frac{1}{x}\int f(x)$ $\frac{1}{1-x} = \int \frac{1}{x}\int f(x)$ $\frac{1}{(1-x)^2} = \frac{1}{x}\int f(x)$ $\frac{x+1}{(1-x)...


1

see Help with summation: $\sum_{k=1}^\infty\frac{k(k+2)}{15^k}$ Reusing that post result: $s_0 = \frac{x}{1-x}, s_1 = \frac{x}{(1-x)^2}$ $$\sum_{n=0}^\infty (2n+1) x^n = 1 + \sum_{n=1}^\infty (2n+1) x^n = 1 + 2 s_1 + s_0 = \frac{1+x}{(1-x)^2}$$


1

$\displaystyle \sum_{n=0}^{\infty}(2n+1)x^{n}$ $\displaystyle = \sum_{n=0}^{\infty}[(2n+2)-1]x^{n}$ $\displaystyle = 2\sum_{n=0}^{\infty}(n+1)x^{n} - \sum_{n=0}^{\infty}x^{n}$ $\displaystyle = 2\sum_{n=0}^{\infty} \frac{\mathrm{d}}{\mathrm{d}x}x^{n+1} - \frac{1}{1-x}$ $\displaystyle = 2 \frac{\mathrm{d}}{\mathrm{d}x}\sum_{n=0}^{\infty} x^{n+1} - \frac{...


2

Hint: $$\sum_{n=0}^{\infty} (n+1)^2 x^n=\sum_{n=0}^{\infty} n(n+1) x^n+\sum_{n=0}^{\infty} (n+1) x^n=x\sum_{n=0}^{\infty}(x^{n+1})''+\sum_{n=0}^{\infty} (x^{n+1})'.$$


2

Hint: Let $f(x)=\sum x^{n}$. Then $f'(x)=\sum nx^{n-1}=\frac 1 x \sum nx^{n}$. Also $f''(x)=\sum n(n-1)x^{n-2}=\frac 1 {x^{2}}\sum n^{2}x^{n}-\frac 1 {x^{2}}\sum nx^{n}$. Use these equations to write down $\sum nx^{n}$ and $\sum n^{2}x^{n}$ in terms of $f(x)$ (which can be computed explicitly). Can you finish now?


1

HINT To begin with, notice that \begin{align*} \sum_{n=0}^{\infty}(2n+1)x^{n} = \sum_{n=0}^{\infty}[(2n+2)-1]x^{n} = 2\sum_{n=0}^{\infty}(n+1)x^{n} - \sum_{n=0}^{\infty}x^{n} \end{align*} Then you make use of the fact that \begin{align*} \begin{cases} \displaystyle\sum_{n=0}^{\infty}x^{n} = \frac{1}{1-x} & \text{for}\,\,|x| < 1\\\\ \displaystyle (n+...


4

Every point on the limiting curve has a distance 1 to the $x$-axis along the tangent line at that point. Intuitively, this should be plausible. More formally, let $x=f(y)$ be the equation for the curve. The equation for the tangent line at $(f(y_0),y_0)$ is $$x-f(y_0)=f'(y_0)(y-y_0).$$ This meets the $x$-axis at $(y_0f'(y_0)+f(y_0),0)$ The condition then ...


0

Use geometric series. The limit of the geometric series with ratio $q$ is $\frac{1}{1 - q}$. Compare this with $\frac{1}{2 - x}$ which will lead you to $q = x - 1$. Then one has $$f(x) = \frac{1}{2 - x} = \frac{1}{1 - q} = \sum_{k = 0}^\infty q^k = \sum_{k = 0}^\infty (x - 1)^k.$$ The Taylor series is centered at $x_0 = 1$ and its radius of convergence is ...


0

Trivially, it is false. For example, let $a_n = b_n = (-1)^n$. Then both $(a_n)$ and $(b_n)$ are divergent sequences. However $a_nb_n=1$, so $(a_n b_n)$ is a convergent sequence.


6

Try to find an example where the two sequences are divergent but the product is constant. Remember that divergent just means that it doesn't converge, it doesn't have to go to infinity in either direction.


2

This is mainly a re-post of my comment: In this question, I have defined $$A_N:=\sqrt{-N}\cdot\frac{E_2(\tau_N)-\frac{3}{\pi\cdot Im(\tau_N)}}{\eta^4(\tau_N)}$$ where $\eta$ denotes the Dedekind $\eta$-Function and $E_2$ is the Eisenstein series of weight $2$, and $\tau_N=\frac{N+\sqrt{-N}}{2}$ is a quadratic irrationality with class number $1$. For the ...


2

Basically the author of the second solution did the exact same thing that you did: They realized that $f_n$ depends on $n$ and $x$ not really separately, but only via their product: $$f_n(x) = f(nx),$$ where $f(x)=e^{-x}\sin(x)$. That means that for any constant $c$ you get $f_n(\frac{c}n)=f(c)$, so the function sequence $f_n$ cannot converge to the null ...


1

Yes, you can clearly apply the integral test here. You can also do without it: Let $$ \forall n \ge2,\, u_n = \frac{5}{n^2\ln(n)} \quad \text{ and } \quad v_n = \frac{1}{n^2} $$ We have $$ \forall n \ge 2,\, \frac{u_n}{v_n} = \frac{5}{\ln(n)} $$ Thus $$ \lim_{n \rightarrow +\infty} \frac{u_n}{v_n} = 0 $$ So $ u_n = o(v_n) $ and since $(u_n)$ and $(v_n)$ are ...


2

The author did not deduce that $\sup\lvert f_n\rvert=e^{-1}\sin(1)$. What he or she deduced was that $\sup\lvert f_n\rvert\geqslant e^{-1}\sin(1)$. It follows from this that we don't have $\lim_{n\to\infty}\sup\lvert f_n\rvert=0$ and that therefore $(f_n)_{n\in\mathbb N}$ doesn't converge uniformly to the null function. But it converges pointwise to that ...


3

Assuming that you meant $$\sum_{k=2}^\infty\frac5{k^2\ln{(k)}}$$ then of course we can use the integral test as stated. We have that $$0\lt\int_3^\infty\frac5{x^2\ln{(x)}}\mathrm{d}x\lt\int_3^\infty\frac5{x^2}\mathrm{d}x=\frac53$$ and hence the sum is convergent. (I use the lower bound $3$ because we require $\ln{(x)}\gt1$ for the inequality to hold)


1

Oops. Read two inequalities backwards. Below there's an example such that $(\epsilon_1+\dots+\epsilon_n)/n$ does not tend to $0$; that answers what seems to me to be an interesting question, but not the question that was actually asked. For $k=0,1,\dots$ let $$A_k=\{n:3^{2k}\le n<3^{2k+1}\}$$and $$B_k=\{n:3^{2k+1}\le n<3^{2k+2}\}.$$Note that the ...


0

I would just like to give a short proof that there is at most one value of $a$ such that the sequence $a_n$ of the OP consists of positive real numbers. Suppose that $a_n,a_n'$ were two such sequences related to two different values $a,a'.$ Then $d_n=a_n-a_n'$ satisfy $d_{n+1}=-\left(\frac n{a_na_n'}+1\right)d_n-d_{n-1}$ for $n\geq1$ and $d_0=0$. Now we ...


3

Yes, it converges uniformly. No, the M-test will not work - if you could find those $C_n$ that would show the series converges absolutely, which is not so. But: Define $$s_n(x)=\sum_{j=1}^n\frac{\cos(j+x)}{j},$$ $$t_n=\sum_{j=1}^n\frac{\cos(j)}{j},$$ $$r_n=\sum_{j=1}^n\frac{\sin(j)}{j}.$$Then $$s_n(x)-s_m(x)=\cos(x)(t_n-t_m)-\sin(x)(r_n-r_m),$$so $$|s_n(x)-...


3

Hint: You can fix $k \in \mathbb{N}^*$ and show by induction on $n$ $$ \mathcal{P}_n : \frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}+\dots +\frac{1}{(n+k-1)(n+k)}= \frac{n}{k(n+k)} $$ To show $\mathcal{P}_n \implies \mathcal{P}_{n+1}$, you have to prove that $$ \frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}+\dots +\frac{1}{(n+k-1)(n+k)} + \frac{1}{(n+k)(n+k+1)}= \frac{n+1}{k(...


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