New answers tagged

1

A straightforward approach is to see how $a_n$ and $b_n$ change when $n$ is increased by $1$. The constant term of $f_1(x)$ is $4$. Suppose that $f_n(x)$ for some $n$; then $f_n(x)=xg_n(x)+4$ for some polynomial $g_n(x)$ (why?), so $$f_{n+1}(x)=\left(\big(xg_n(x)+4\big)-2\right)^2=\big(xg_n(x)+2\big)^2=x^2\big(g_n(x)\big)^2+4xg_n(x)+4\;.$$ The first two ...


0

Consider the sequence $(p_n)=\{-9,9,0,0,0,0,0,\dots\}$. Certainly $\liminf p_n=\limsup p_n=0$, so $-1<\liminf p_n=\limsup p_n<1$. However there is no $0<P<1$ such that $|9|\leq P$. This seems like cheating though. Do you mean to have $0<P$?


1

First note that $$\sum_{b=0}^ax^b=\frac{1-x^{a+1}}{1-x}$$ Now, $$\sum_{a=1}^{s+t-1}\sum_{b=0}^ax^b=\sum_{a=1}^{s+t-1}\frac{1-x^{a+1}}{1-x}=\frac{1}{1-x}\sum_{a=1}^{s+t-1}\left(1-x^{a+1}\right)=$$ $$=\frac{1}{1-x}\left(\sum_{a=1}^{s+t-1}1-\sum_{a=1}^{s+t-1}x^{a+1}\right)=\frac{1}{1-x}\left(s+t-1+x-\sum_{a=1}^{s+t}x^a\right)$$ The second one can be calculated ...


1

There is no such sequence. Hint: Assume otherwise, that $\left\{P_n\right\}$ is a sequence as described. Consider $P_n=1-\frac{1}{n}$, and prove that there is a subsequence of $\left\{p_n\right\}$ which converges to $1$.


1

One way to tackle this is to do induction on $m$. Assume that $m=1$, then $$|a(1)-x|=\sum_{j=1}^n|b(j)-x|$$ for all $x\in\mathbb R$. Thus by (reverse) triangle inequality, $$\sum_{j=2}^n|b(j)-x|=|a(1)-x|-|b(1)-x|\le ||a(1)-x|-|b(1)-x||$$ $$\le |(a(1)-x)-(b(1)-x)|=|a(1)-b(1)|.$$ Suppose for a contradiction that $n>1$ so that $a(1)\ne b(1)$, then by ...


4

Assume by contradiction that $A$ is unbounded. Let $a \in A$. Then for each $n$ there exists some $x_n$ such that $$d(x_n,a) >n$$ Now, $x_n$ has a converging subsequence $x_{k_n} \to b$. Now, for $\epsilon=1$, since $x_{k_n} \to b$ there exists some $N$ so that for all $n >N$ we have $$d(x_{k_n}, b) <1$$ Then, for all $n >N$ we have $$d(a,b) ...


0

$$\zeta(1)=\sum_{n\ge1}^{\Re}\frac1n=\lim_{N\to\infty}\left(\sum_{n=1}^N\frac1n-\int_1^N\frac1tdt\right)=\gamma$$ It equals the Euler-Mascheroni constant by definition. You can find the said formula and derivation on the wikipedia.


0

First show this is true if $c_n = 1$ for all $n$. Separately, show that if $\{d_n\}$ is another sequence with $d_n \to 0$ then $\frac 1 M \sum_1^M d_n a_n \to 0$. (Use the fact that $\{a_n\}$ is bounded and for any $\epsilon > 0$, all except a finite number of the $d_n$ satisfy $|d_n|<\epsilon$.) Now put these two observations together and you are ...


0

The given inequality is equivalent with $$1+\cdots +\frac{1}{n-1}-\ln n < \gamma$$ It now follows by remarking that $$b_n=1+\cdots +\frac{1}{n}-\ln (n+1)$$ (which has the same limit $\gamma$ as $a_n$) is increasing. This being easy to see from the inequality $$\ln (1+\frac{1}{n})\leq \frac{1}{n}$$


0

$\displaystyle (\lim a_n) -a_n = \sum_{k=n}^\infty a_{k+1}-a_k = \sum_{k=n}^\infty \frac{1}{k+1} - \ln\left(1-\frac{1}{k+1} \right)$ Since $\forall x> 0, x-\frac{x^2}2\leq \ln(1+x)< x$, $$-\frac 12 \sum_{k=n}^\infty \frac{1}{(k+1)^2} \leq (\lim a_n) -a_n < 0$$ A standard integral estimate yields $\displaystyle \frac 12 \sum_{k=n}^\infty ...


0

Yes, this was a typo. Thanks for comments. Trying to close.


2

Denote $\sum a_n x^n$ the given series then $$\left\vert\frac{a_{n+1}}{a_n}\right\vert=\left(1+\frac1n\right)^{-n}\xrightarrow{n\to\infty}\frac1e$$ so by the ratio test, the radius of convergence is $e$.


0

Note that $$\sum_{k=1}^\infty x^k = \frac{1}{1-x} - 1$$ converges uniformly on every compact subset of $(-1,1)$ so that $$\sum_{k=1}^\infty k x^{k-1} = \frac{1}{(1-x)^2}$$ by differentiation and $$\sum_{k=1}^\infty kx^k = \frac{x}{(1-x)^2}$$ after multiplication by $x$. Repeat five times and plug in $x=\frac 12$.


0

What do you mean by "period of the graph"? You wouldn't really use the graph, but rather the basic trigonometric identity $\sin^2(x)+\cos^2(x) = 1$ or the very definition of sine.


6

Note that $$\frac{1}{n\left(n-1\right)}=\frac{1}{n-1}-\frac{1}{n}$$ What does it tell you?


1

Iduction should work well: Assume we already know that $a_n<1+3^{n-1}$ and $a_{n+1}<1+3^n$. Then $$ \begin{align}a_{n+2}&=a_{n+1}+5a_n\\&<1+3^n+5\cdot(1+3^{n-1})\\&=6+8\cdot 3^{n-1}\\&=6+3^{n-1}-3^{n-1}\\ &<1+3^{n+1}\end{align}$$ provided that $3^{n-1}>5$. This last requirement holds as soon as $n\ge 3$, so you need to ...


2

Induction, perhaps. For $\;n=1,2\;$ it is trivial, so suppose it is true for all $\;k<n+2\;$ and let us try to prove for $\;k=n+2\;,\;\; n\ge3\;$ : $$a_{n+2}=a_{n+1}+5a_n\stackrel{\text{Ind. Hyp.}}<1+3^n+5+5\cdot3^{n-1}=6+8\cdot3^{n-1}$$ Now, we'd like to show $$6+8\cdot3^{n-1}\le1+3^{n+1}\iff1+3^{n-1}(9-8)\ge6$$ and it's easy to check the last ...


1

$$\sum_{n\geq 1}\frac{3n-1}{(n+1)^3} = 3\sum_{n\geq 1}\frac{1}{(n+1)^2}-4\sum_{n\geq 1}\frac{1}{(n+1)^3} = 3(\zeta(2)-1)-4(\zeta(3)-1) = \color{red}{3\zeta(2)-4\zeta(3)+1}.$$


0

$$\frac{g}{(1+u)^g-1}-\frac{1}{u}=\frac{gu-(1+u)^g+1}{u((1+u)^g-1)}=\frac{1+gu-(1+u)^g}{u((1+u)^g-1)}$$ $$=\frac{\sum_{k=2}^g \binom{g}{k}u^k}{\sum_{k=1}^g\binom{g}{k}u^{k+1}}\approx\frac{\binom{g}{2}u^2}{\binom{g}{1}u^2}=\frac{g-1}{2} \text{ as } u\to0$$ $$\implies G(u)=\frac{\sum_{k=2}^g ...


8

$$ \sum_{n=1}^\infty\frac{3n-1}{(n+1)^3}\le\sum_{n=1}^\infty\frac{3n}{(n+1)^3}\le\sum_{n=1}^\infty\frac{3n}{n^3}=3\sum_{n=1}^\infty\frac1{n^2}<\infty. $$


3

Writing $x_n = \frac{1-p^{n+1}}{1-p^n}\frac{n}{n+1}$, we see: $$x_n>x_{n-1}\iff \frac{1-p^{n+1}}{1-p^n}\frac{n}{n+1}>\frac{1-p^{n}}{1-p^{n-1}}\frac{n-1}{n}$$ $$\iff(1-p^{n-1})(1-p^{n+1})n^2>(1-p^n)^2(n^2-1)$$ $$\iff (1-p^n)^2 > n^2[(1-p^n)^2-(1-p^{n-1})(1-p^{n+1})]=n^2[p^{n-1}+p^{n+1}-2p^n]$$ ...


0

The idea behind the proof is quite simple once you get read of the useless verbosity. You start by using the fact that $\{n: |u_n-t|<1\}$ is infinite, take $n_1$ in it. Then you can use the fact that $\{n: |u_n-t|<\dfrac12\}$ is infinite, take $n_2$ in it strictly bigger than $n_1$. Proceeding as this by induction, for each $k$ you can choose $n_k$ ...


2

If you look at the left-hadn side of the equals sign, you have $$ \begin{align} \frac{g}{gu + \cdots + u^g} - \frac{1}{u} &= \frac{1}{u}\frac{g}{g + {g \choose 2} u + \cdots + u^{g-1}} - \frac{1}{u} \\ &= \frac{1}{u}\frac{1}{1 + \frac{1}{g}{g \choose 2} u + \cdots + \frac{1}{g}u^{g-1}} - \frac{1}{u} \end{align} $$ If you then expand the first term ...


2

One may recall that, as $x \to 0$, by the Taylor series expansion, we have $$ \frac1{1+x}=1-x+O(x^2). $$ We assume $g\neq0$ and $ u \to 0$. Then we have $$ \begin{align} \frac{g}{gu+\cdots+u^g}&=\frac1{u+u\left(1/g+\cdots+u^{g}/g\right)} \\\\&=\frac1{u}\frac1{1+u\left(1/g+\cdots+u^{g-1}/g\right)} ...


1

I'm going to assume that in your question the indeces in the summation are from $n=0$ to $\infty$ instead of $j=0$ to $\infty$; if they are not then both series diverge, as $$ \sum_{j=0}^{\infty} z^n = \lim_{m \to \infty} \sum_{j=0}^{m} z^n = \lim_{m \to \infty} mz^n = \infty. $$ The basic idea for the Weierstrass $M$-test is that you bound the norm of your ...


0

$b$ is not be 0. $a$ and $b$ are real numbers. $x_n=a+\alpha$, where $\alpha \to 0$, when $n \to \infty$. $y_n=\frac{x_n}{b+\beta}$, where $\beta \to 0$, when $n \to \infty$. $y_n=\frac{a+\alpha}{b+\beta}$. $y_n \to \frac{a}{b}$.


2

HINT: $$\begin{align} \left|y_n -\frac ab\right|&=\left|\frac{x_n}{\frac{x_n}{y_n}} -\frac ab\right|\\\\ &=\left|\frac{b(x_n-a)-a\left(\frac{x_n}{y_n}-b\right)}{b\frac{x_n}{y_n}}\right| \end{align}$$ Now, apply the triangle inequality.


1

For a linear recurrence relation with constant coefficients, the basic candidates for solutions are $r^n$ where $r$ is any root of the characteristic polynomial. For instance in your example the characteristic polynomial is $p(x)=x^2-6x+9$ which has a root of $3$. Any linear combination of these solutions is also a solution because of linearity. It can ...


1

Hint: Write $$a = (a_n + b_n)a.$$ Then evaluate the difference: \begin{align*} & |a_nz_n + b_nt_n - a| \\ = & |a_nz_n + b_nt_n - (a_n + b_n)a| \\ = & |a_n(z_n - a) + b_n(t_n - a)| \\ \leq & |a_n||z_n - a| + |b_n||t_n - a| \\ \leq & \cdots. \end{align*}


0

The expression $f(z)=\frac1{z-3}$ is already the Laurent series around $|z-3|>0$.


1

Maybe you are looking for this: $(-1)^nx^{2n}$. One may recall that $$ \cos x= \sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}x^{2n} $$ giving $$ f(x)=\cos x. $$


1

$$\lim a_n = L \implies \lim a_n-L=0 \implies \forall\epsilon>0, \;\exists N_{\epsilon} \in \mathbb{N} : |a_n-L|\leq \epsilon,\; \forall n>N_{\epsilon}$$ $$\implies (a_n-L)^2<\epsilon^2 \implies a_n^2-2a_nL+L^2 < \epsilon^2,\; \forall n>N_{\epsilon}$$ So, $$\lim a_n^2-2a_nL+L^2 = 0$$ However, $$\lim a_nL = L^2$$ as limits are preserved ...


2

You can combine techniques. For example $\sum b_n$ where $b_n=2^n/3^n$ converges as a geometric series. On the other hand, $n/3^n < b_n$, so your sum converges by the comparison test.


3

You are close. For a start, you can assume that $\epsilon_1 < 1$. This means that $|a_n^2 -L^2| <\epsilon_1|a_n+L| <\epsilon_1(2L+\epsilon_1) <\epsilon_1(2L+1) $. Then, to make $\epsilon_1(2L+1) < \epsilon $, just choose $\epsilon_1 <\dfrac{\epsilon}{2L+1} $.


0

You can use, more generally, the fact that continuous functions and limits commute. That is, if $f$ is continuous, and $(a_{n})_{n}$ converges to $L$, $f(a_{n})\rightarrow f(L)$. (take $f(a)=a^2$). Proof here.


1

In case this was a question in good faith: since this problem is equivalent to the Riemann Hypothesis, don't expect to actually prove it. LAGARIAS For the curious, the best of the elementary RH versions to experiment with is the first one, by Jean-Louis Nicolas (1983). It says that RH is equivalent to the conjecture that, for all primorial numbers $P,$ ...


5

$\frac{1}{(1-x)(1-x^3)}=\frac{1}{x-x^3}(\frac{1}{1-x}-\frac{1}{1-x^3})$ $\frac{x^2}{(1-x)(1-x^5)}=\frac{1}{x-x^3}(\frac{1}{1-x^3}-\frac{1}{1-x^5})$ ... So, assuming n starts from 1, $\frac{1}{(1-x)(1-x^3)}+\frac{x^2}{(1-x^3)(1-x^5)}+\frac{x^4}{(1-x^5)(1-x^7)}...=\frac{1}{x-x^3}(\frac{1}{1-x}-\frac{1}{1-x^{2n+1}})$


6

Hint: $$\frac{x^{2n}}{(1-x^{2n+1})(1-x^{2n+3})}=\frac{1}{x(1-x^2)}(\frac{1}{1-x^{2n+1}}-\frac{1}{1-x^{2n+3}})$$


0

HINTS: We show uniform convergence of $f_n(x) \to 0$ on $(0,1)$ by demonstrating: $$\forall \epsilon > 0, \exists N\in \mathbb{N}:\forall x\in(0,1), \left| \dfrac{x}{1+nx}\right|<\epsilon \; \forall n>N$$ This reduces to the following problem Given an $x\in(0,1)$ and an $\epsilon > 0$ can you derive a fomula for a value of $N$ that makes ...


0

Assume $n>0$ for the recursion. $$t_1=2\cdot 1\cdot t_0$$ $$t_2=2\cdot 2\cdot t_1=2\cdot 2\cdot 2\cdot 1\cdot t_0 $$ $$t_3=2\cdot 3 \cdot t_2=2\cdot 3 \cdot 2\cdot 2\cdot 2\cdot 1\cdot t_0$$ The pattern should be clearer this way, as for each substitution you get a additional power for $2$ and you get the product of integer numbers (factorial). In ...


1

Your sequence is decreasing, so you should provide a lower bound not a upper bound. The rest part is correct I think. Just a hint: $$2x_n+\frac{a}{x_n^2}=x_n+x_n+\frac{a}{x_n^2}$$ then you can apply the inequality of arithmetic and geometric means, if $a\geq 0$.


1

You're looking at $$ \sum_{n=0}^{\infty}\left(\frac{n x}{(n!)^{2/n}}\right)^n. $$ If the expression in parentheses ever drops below $1$ and stays there, then the series converges geometrically. If you know that $(n!)^{1/n}\sim n/e$, then you can see that this will eventually happen for any $x$, since the expression in parentheses is asymptotic to ...


0

Yet another approach, using only elementary facts: (written here for the sake of getting an alternative proof) You can write $\lvert x\rvert n \leq x^2 \mathbb{1}_{\{\lvert x\rvert > n\}} + n^2 \mathbb{1}_{\{\lvert x\rvert \leq n\}}$ so that $$ 0 \leq \sum_{n=0}^\infty \frac{(\lvert x\rvert n)^n}{n!^2} \leq \sum_{n=0}^\infty ...


1

To say that $\displaystyle \limsup_{n \to \infty} \frac{a_n}{b_n} = L_2$ means: first, for any positive $\epsilon > 0$ there exists $N$ with the property that $n \ge N$ implies $\dfrac{a_n}{b_n} < L_2 + \epsilon$, and second, $L_2$ is the least number with this property. This is more than is required. It suffices to know that for some $N$, ...


2

If we try the ratio test, $$ \frac{\frac{(n+1)^{n+1}x^{n+1}}{((n+1)!)^2}}{\frac{n^nx^n}{(n!)^2}} =\frac{(n+1)^{n+1}x}{n^n(n+1)^2}=\left(1+\frac1n\right)^n\frac{x}{n+1}\to0. $$ As $n\to\infty$. So the series converges for all $x$.


2

Denote $\sum a_n x^n$ this series. We have $$\frac{a_{n+1}}{a_n}=\left(1+\frac1n\right)^n\frac1{n+1}\xrightarrow{n\to\infty} e\times0=0$$ so by the ratio test the radius of convergence is $R=\frac10=+\infty$.


0

You will get a few answers that use the Sterling formula for the asymptotic behavior of $n!$. You don't need that though. Observe that 2/3 of the factors in $n!$ are greater than $n/3$, and therefore $n! > (n/3) ^ {(2n/3)}$. Use this in your computation of the radius of convergence for your series.


2

You can prove by induction that : $0 \leqslant u_n \leqslant \frac{2}{n(n+1)}u_1$. Base case for $n=1$, that is okay Inductive step Assume it is true for all $k < n$. Then : $u_n \leqslant \frac{u_n}{n^2} + \frac{1}{n^2}\sum \limits_{k=1}^{n-1} u_k$, so $(1 - \frac{1}{n^2})u_n \leqslant \frac{1}{n^2}\sum \limits_{k=1}^{n-1} \frac{2}{k(k+1)}u_1 = ...


0

Can we apply the root test ? The nth term and (n+1)th term of the series is $|T_{n}|=\frac{\sqrt{a_{n}}}{(n)^{p}}$ and $|T_{n+1}|=\frac{\sqrt{a_{n+1}}}{(n+1)^{p}}$ respectively. Then by root test $|\frac{T_{n+1}}{T_n}|= \sqrt{\frac{a_{n+1}}{a_{n}}} \frac{n^p}{(n+1)^p}$ , where the series $\sum_{i=1}^\infty a_{n}$ is convergent , so $\frac{a_{n+1}}{a_{n}} ...


2

We have $\vert\sin x\vert \le \vert x\vert$ and to conclude it suffices to prove that the series $\sum \vert x\vert^{\sqrt n}$ is convergent. We have $$n^2 \vert x\vert^{\sqrt n}=\exp(\sqrt n\ln\vert x\vert+2\ln n)\xrightarrow{n\to\infty}0$$ so we have $\vert x\vert^{\sqrt n}\le \frac1{n^2}$ for $n$ large enough and the result follows by comparison with a ...



Top 50 recent answers are included