Articles of sequences and series

Show that $q^4+2pq^2 +p^2 = 2pq -(pq)^2 -1$ becomes $p^3+q^3+3pq-1=0$.

I know that these two are exactly the same equation but I can’t seem to prove it. You are also given that $p+q=1$. This is a follow up from a similar question.

Find two constant$c_1, c_2$ to make the inequality hold true for all N

Show that the following inequality holds for all integers $N\geq 1$ $\left|\sum_{n=1}^N\frac{1}{\sqrt{n}}-2\sqrt{N}-c_1\right|\leq\frac{c_2}{\sqrt{N}}$ where $c_1,c_2$ are some constants. It’s obvious that by dividing $\sqrt{N}$, we can get $\left|\frac{1}{N}\sum_{n=1}^N \frac{1}{\sqrt{n/N}} -2 – \frac{c_1}{\sqrt{N}}\right|\leq \frac{c_2}{N}$ the first 2 terms in LHS is a Riemann Sum of $\int_{0}^1\frac{1}{\sqrt{x}}dx$. But what is the next step? I have not seen a […]

Convergence of $\sum_{n=1}^{\infty} a^{-\frac{1}{c+1}}\prod_{k=1}^{n} \frac{a}{a+k^{c}}$ as $a\to\infty$

It looks to me, by doing numerical simulations, that $$ f(c) = \lim_{a\rightarrow \infty}a^{-\frac{1}{c+1}}\sum_{n=1}^{\infty} \prod_{k=1}^{n} \frac{a}{a+k^{c}} $$ converges and that $1\leq f(c) <2$, which was quite surprizing to me. It is easy to see that $f(0)=1$ (using geometric series) and it has been shown that $f(1)= \sqrt{\frac{\pi}{2}}$, as shown by the answer to Convergence of […]

Convergence of the series $\sum_{n=1}^{\infty}\frac{3\ln(n)}{n^7}$; I say it converges, WebWork tells me this is incorrect.

I’m almost certain that $$\sum_{n=1}^{\infty}\frac{3\ln(n)}{n^7}$$ converges. However, WebWork tells me that this is incorrect. I have been in that situation before but I obviously can’t assume that I’m right and the computer is wrong based on that. I don’t know how the system work and I’m not sure whether what I do is correct anymore. […]

Next number in series

What are the basic/advanced strategies used to find the next number in series. I know the simple ones such as addition, multiplication etc. But recently I came into a question that goes on something like 812, 819, 823, 835, 834, 851(Don’t try to solve this, I changed some numbers and there is no sequence). This […]

If $\sum a_n^2 $ converges, show that $\lim_{n\to\infty}b_n=0$

Let $(a_n)_n \subset R$ and let $b_n=\dfrac{a_1+a_2+…+a_n}{n}$. Prove that if $\sum_{n=1}^{\infty} a_n^2 $ converges, then $\lim\limits_{n\to\infty} b_n=0$. From the hypothesis, $\sum_{n=1}^{\infty} a_n^2 $ converges, we get $\lim a_n=0$. Then I have no idea for this problem. Can anyone help me please? Thanks in advanced.

Branch cut for $\sqrt{1-z^{2}}$ and Taylor's expansion!

I’m working in a problem that involves the equation $$ w(z)=\sqrt{1-z^{2}} \,\, . $$ I already know that there’re two branch points in this equation, namely $\pm 1$, so there’s a Riemann surface covering the domain of the function where the branch cut is from the $-1$ to $1$, as shown in the figure below. […]

Approximating a polynomial to a piece-wise function

I was going through my introductory calculus book(for high-school student) by a Russian author(N.Psikunov) where I encountered a theorem without proof named: Weierstrass Approximation Theorem So how can we apply this theorem and apply it to piece wise functions?(any general approach?)(say a simple function like |x|)

Explicit formula for a sequence

I want to construct an infinite series which goes like this: $\Large x^2+x^{2x^2}+x^{2x^{2x^2}}+x^{2x^{2x^{2x^2}}}+…$ Therefore, I need an explicit formula of the sequence $\Large x^2,x^{2x^2},x^{2x^{2x^2}},x^{2x^{2x^{2x^2}}},…$ I tried this recursive formula first, but it turned out to be incorrect: $(a_{n+1})=(a_n)^{(a_n)}$ Does there even exist an explicit (or at least a recursive) formula for this sequence?

Prove that if $ |a_n-a_{n-1}| < \frac{1}{2^{n+1} }$ and $a_0=\frac12$, then $\{a_n\}$ converges to $0<a<1$

I try to solve this question but I don’t know how. given $ a_0 = \frac12 $ and for each $n\geq 1$: $$ |a_n-a_{n-1}| < \frac{1}{2^{n+1}} $$ show that $\{a_n\}$ converges and the limit is $a$ such that $0<a<1$ Update (Edited): I showed by cauchy that $ |a_m-a_n| < |a_m-a_{m-1}+a_{m+1}-…+a_{n+1}-a_n| < \frac{1}{2^{m+1}} + \frac{1}{2^{m}}+…+\frac{1}{2^{n+2}}$ by […]