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7

We have $$\frac{a_{n}}{a_{n+1}}=\frac{3^n-7}{3^{n+1}-7}\cdot \frac{4^{n+1}+5}{4^n+5}.$$ The ratio $\dfrac{a_n}{a_{n+1}}$ has limit $4/3$ as $n\to\infty$. (Divide top and bottom of the first term on the right by $3^n$, and top and bottom of the second term by $4^n$.) So by the definition of limit there is an $N$ such that if $n\gt N$ then ...


4

Since $$\begin{align*} \left\| \sum_{k=1}^n (X_k-\mathbb{E}(X_k)) - \sum_{k=1}^m (X_k-\mathbb{E}(X_k)) \right\|_{L^2}^2 &= \left\| \sum_{k=m+1}^n (X_k-\mathbb{E}(X_k)) \right\|_{L^2}^2 \\ &= \sum_{k=m+1}^n \sum_{\ell=m+1}^n \text{cov}(X_k,X_{\ell}) \end{align*}$$ for all $n \geq m$, we find by the Cauchy Schwarz inequality $$\begin{align*} \left\| ...


3

Calculate $a_{n+1}-a_n$, then $$\frac{3^{n+1}-7}{4^{n+1}+5}-\frac{3^n-7}{4^n+5}=\frac{-3^n 4^n +21\cdot 4^n + 10\cdot 3^n}{(4^{n+1}+5)(4^n+5)}. $$ If we show $-3^n 4^n +21\cdot 4^n + 10\cdot 3^n < 0$ for large $n$, then the proof is over. Since $3^n < 4^n$ for all $n\in\mathbb{N}$, $$-3^n 4^n +21\cdot 4^n + 10\cdot 3^n < -3^n4^n + 31\cdot ...


2

Hint: Assume $(x_n)$ is C-convergent to $x$. Fix $n > N$. Then for all $\epsilon > 0$ we know that $0 \leq d(x_n,x) < \epsilon$. So $d(x_n,x) \in \cap_{\epsilon > 0} [0,\epsilon) = \{0\}$, which is to say that $x_n = x$. Thus a sequence $(x_n)$ is C-convergent if and only if there exists $x \in \mathbb{R}$ and $N$ such that $x_n = x$ for all $n ...


2

(this proof takes a little longer, but gives a neat bound on the rate of convergence) If you want a universal bound of the form $a_n\le M$, the recursion tells you that you're going to need $M\le M^2+\frac{M}{6}+\frac{1}{9}$, which works out to be equivalent to $(6M-1)(3M-2)\le0$. The lowest bound this allows us to take is $M=\frac{1}{6}$, i.e. that ...


2

This works for sums of $p$th powers of $k$ because of the fact that $(n+1)^{p+1}-n^{p+1},$ when expanded by the binomial theorem, will have no $n^{p+1}$ term, and so when summed only uses powers up to the $p$th power. Also before expanding it, its sum "telescopes" (all terms cancel but two, or all but one if you sum starting at $0.$). Also once you accept ...


1

Here's how you do it! I've attached a picture I found on the internet since typing it takes time. This is a general approach that works for the sum of the $k^{th}$ power of n consecutive integers for all possible values of n. Hope it helps :)


1

You need the the assumption that the series $\sum_n a_n$ is convergent: otherwise, the quantity $\sum_{k=n+1}^\infty a_k$ cannot be manipulated (it is not defined). Furthermore, the inequality that actually holds is in the other direction. See below. Assuming the series $\sum_n a_n$ is convergent, you can write: $$ \sum_{k=n+1}^\infty a_k = ...


1

Let $u=x+A/x$ so that $x^2-u x+A=0$, or $$x = \frac{u}{2} \pm \frac12 \sqrt{u^2-4 A} $$ $$dx = \frac12 \left (1 \pm \frac{u}{\sqrt{u^2-4 A}} \right ) du$$ Note that the integration limits provided by the mapping $x \mapsto u$ depend on whether the point $x=B$ is less than or greater than the minimum of $u$ at $x=\sqrt{A}$. Let's assume the former, i.e., ...


1

Yes. Removing finitely many terms from a sequence does not influence its accumulation points.


1

Your convergence statement is false, because Dirichlet's test assumes monotonicity. Indeed $x_n=(-1)^n \frac{1}{n}$ is in $c_0$ but $\sum_{n=1}^\infty (-1)^n x_n = +\infty$.


1

$C$-convergence is much stronger than convergence. The sentence for every $\epsilon>0$ and every $n>N$, $d(x_n,x)<\epsilon$ is the same as for every $n>N$, $x_n=x$.


1

Differentiating with respect to $n$, you get $(4^n + 5)(3^n \log3 ) - (3^n - 7)(4^n \log4) \implies 12^n \log(0.75) + 5 · 3^n · \log3 + 7 · 4^n · \log4· \log(0.75) < 0$; Therefore, for large $n$, the derivative is less than $0$. Thus, the sequence is decreasing.


1

This is true only if $\lvert z\rvert <1$. The reason is quite simple and relies on a well known identity from high school: $$1-z^n=(1-z)(1+z+z^2+\dots+z^{n-1})$$ which we can rewrite as $$\frac1{1-z}=1+z+z^2+\dots+z^{n-1}+\frac{z^n}{1-z}.$$ From this we deduce that $$\biggl\lvert\frac1{1-z}-(1+z+z^2+\dots+z^{n-1})\biggr\rvert=\frac{\bigl\lvert ...


1

Note that $\lim_{n->\infty}{(a_{n}-b_{n})}$ can be $0$ even if $a_n$ and $b_n$ do not converge. For instance take $a_n=b_n=(-1)^n.$


1

You made a mistake writing the ratio. It should be $$ \frac{e^{n+6}}{\sqrt{n+8}(n+4)!} \frac{\sqrt{n+7}(n+3)!}{e^{n+5}} = \frac{e\cdot e^{n+5}}{\sqrt{n+8}(n+4)(n+3)!} \frac{\sqrt{n+7}(n+3)!}{e^{n+5}} $$ not $$\frac{e^{n+6}}{\sqrt{n+8}(n+4)n!} \frac{\sqrt{n+7}(n+3)n!}{e^{n+5}}.$$ The mistake is in that you wrote $(n+k)! = (n+k)n!$, which is not true.


1

Note that $$\dfrac{a_{n+1}}{a_n}=\dfrac{\dfrac{e^{n+6}}{(n+4)!\sqrt{n+8}}}{\dfrac{e^{n+5}}{(n+3)!\sqrt{n+7}}}=\dfrac{e\sqrt{n+7}}{(n+4)\sqrt{n+8}}\underbrace{\to}_{n\to \infty} 0.$$



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