Articles of sequences and series

Sum of real powers: $\sum_{i=1}^{N}{x_i^{\beta}} \leq \left(\sum_{i=1}^{N}{x_i}\right)^{\beta}$

Let $\{x_i\}_{i=1}^{N}$ be positive real numbers and $\beta \in \mathbb{R}$. Can we say that: $$ \sum_{i=1}^{N}{x_i^{\beta}} \leq \left(\sum_{i=1}^{N}{x_i}\right)^{\beta}$$ I know that this holds if $\beta \in \mathbb{N}$. Does the above inequality have a name in case it’s true?

How to bound the truncation error for a Taylor polynomial approximation of $\tan(x)$

I am playing with Taylor series! I want to go beyond the basic text book examples ($\sin(x)$, $\cos(x)$, $\exp(x)$, $\ln(x)$, etc.) and try something different to improve my understanding. So I decided to write a program for approximating $\tan(x)$. But I am having difficulty. I want to use the Taylor series of $\tan(x)$ to approximate […]

Limit of $a_n = \sum\limits_{k=1}^{n} \left(\sqrt{1+\frac{k}{n^2}}-1\right)$

Given $a_n = \sum\limits_{k=1}^{n} \left(\sqrt{1+\frac{k}{n^2}}-1\right)$, find $\lim\limits_{n \to \infty} a_n$. My try: To simplify, $$a_n = \frac{\displaystyle\sum_{k=1}^{n}\sqrt{n^2+k}-n}{n}$$ and I’m stuck from there. In addition, I have made a program to find the limit, which says it’s 1/4. Can anybody give me a hint to start? Thanks for your time!

Closed form of $ x^3\sum_{k=0}^{\infty} \frac{(-x^4)^k}{(4k+3)(1+2k)!} $

I was trying to find a closed form of $$ f(x) = x^3\sum_{k=0}^{\infty} \frac{(-x^4)^k}{(4k+3)(2k+1)!} $$ $f(x)$ converges for all $x$ by the ratio test. I began by making the it look like a power series and then differentiating $$ f(x) = \sum_{k=0}^{\infty} \frac{(-1)^kx^{4k+3}}{(4k+3)(2k+1)!} $$ $$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \sum_{k=0}^{\infty} \frac{(-1)^kx^{4k+3}}{(4k+3)(2k+1)!} $$ $$ \frac{\partial […]

Easy way to find out limit of $a_n = \left (1+\frac{1}{n^2} \right )^n$ for $n \rightarrow \infty$?

What’s an easy way to find out the limit of $a_n = \left (1+\frac{1}{n^2} \right )^n$ for $n \rightarrow \infty$? I don’t think binomial expansion like with $\left (1-\frac{1}{n^2} \right )^n = \left (1+\frac{1}{n} \right )^n \cdot \left (1-\frac{1}{n} \right )^n$ is possible. And Bernoulli’s inequality only shows $\left (1+\frac{1}{n^2} \right )^n \geq 1 + […]

$C$ such that $\sum_{k\in \mathbb{Z}^n} k_i^2k_j^2|a_{ij}|^2 \leq C \sum_{k\in \mathbb{Z}^n} \|k\|^4|a_{ij}|^2$

More generally, can we find $C_n>0$ such that $$\sum_{k\in \mathbb{Z}^n} k_i^2k_j^2|a_{ij}|^2 \leq C \sum_{k\in \mathbb{Z}^n} \|k\|^2|a_{ij}|^4$$ for all $\{a_k\}_{k\in \mathbb{Z}^n} \in \ell^2(\mathbb{Z}^n)$ where the above sums converge? My ideas so far: $ij \leq i^2 + j^2$ shows that $C=1$ works. Is 1 the sharpest possible bound? (This comes from showing that for $u,f$ periodic and […]

Need help to prove

I got the result below during my research. $$1=\frac{1}{1+a_1}+\frac{a_1}{(1+a_1)(1+a_2)}+\frac{a_1a_2}{(1+a_1)(1+a_2)(1+a_3)}+\frac{a_1a_2a_3}{(1+a_1)(1+a_2)(1+a_3)(1+a_4)}+… \tag 1$$ $$1=\frac{1}{1+a_1}+\sum\limits_{k=1}^ \infty\frac{\prod\limits_{j=1}^{k}a_j}{\prod\limits_{j=1}^{k+1} (1+a_j)} $$ $$1+a_1=1+\frac{a_1}{1+a_2}+\frac{a_1a_2}{(1+a_2)(1+a_3)}+\frac{a_1a_2a_3}{(1+a_2)(1+a_3)(1+a_4)}+…$$ $$a_1=\frac{a_1}{1+a_2}+\frac{a_1a_2}{(1+a_2)(1+a_3)}+\frac{a_1a_2a_3}{(1+a_2)(1+a_3)(1+a_4)}+…$$ We can get the same relation but without $a_1$ $$1=\frac{1}{1+a_2}+\frac{a_2}{(1+a_2)(1+a_3)}+\frac{a_2a_3}{(1+a_2)(1+a_3)(1+a_4)}+…$$ Examples: Example-1: $a_n=c$ $$1=\frac{1}{1+c}+\frac{c}{(1+c)^2}+\frac{c^2}{(1+c)^3}+\frac{c^3}{(1+c)^4}+…$$ $$1+c=1+\frac{c}{1+c}+\frac{c^2}{(1+c)^2}+\frac{c^3}{(1+c)^3}+…$$ We know that if $\frac{c}{(1+c)}<1$ then $$1+\frac{c}{1+c}+\frac{c^2}{(1+c)^2}+\frac{c^3}{(1+c)^3}+…=\frac{1}{1-\frac{c}{(1+c)}}=1+c$$ Example-2: $a_n=x^{2^{n-1}}$ $$1=\frac{1}{1+x}+\frac{x}{(1+x)(1+x^2)}+ \frac{x^3}{(1+x)(1+x^2)(1+x^4)}+ \frac{x^7}{(1+x)(1+x^2)(1+x^4)(1+x^8)}+…$$ $$1=\frac{1-x}{1-x^2}+\frac{x(1-x)}{1-x^4}+\frac{x^3(1-x)}{1-x^8} +\frac{x^7(1-x)}{1-x^{16}} +…$$ $$1=\frac{1-x}{1-x^2}+\frac{x(1-x)}{1-x^4}+\frac{x^3(1-x)}{1-x^8} +\frac{x^7(1-x)}{1-x^{16}} +…$$ $$\frac{x}{1-x}=\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\frac{x^4}{1-x^8} +\frac{x^8}{1-x^{16}} +…$$ If we […]

Computing $\sum\limits_{n=1}^q {1\over (kn-1)(kn)(kn+1)}$ as a telescoping series, when $k\geqslant3$?

One knows that $$S_1(q)=\sum_{n=1}^q {1\over (n-1)(n)(n+1)} = \sum_{n=1}^q\frac12\left({\frac{1}{(n-1)n}-\frac{1}{n(n+1)}}\right)$$ and the RHS can be easily telescoped. The same approach works for $$S_2(q)=\sum_{n=1}^q {1\over (2n-1)(2n)(2n+1)}$$ However, for $$S_3(q)=\sum_{n=1}^q {1\over (3n-1)(3n)(3n+1)}$$ it is impossible to telescope using the same method than in the two cases above. So: How should $$S_k(q)=\sum_{n=1}^q {1\over (kn-1)(kn)(kn+1)}$$ where $k\geqslant3$ is an integer, be […]

Show that, for all $n > 1: \log \frac{2n + 1}{n} < \frac1n + \frac{1}{n + 1} + \cdots + \frac{1}{2n} < \log \frac{2n}{n – 1}$

I’m learning calculus, specifically limit of sequences and derivatives, and need help with the following exercise: Show that for every $n > 1$, $$\log \frac{2n + 1}{n} < \frac1n + \frac{1}{n + 1} + \cdots + \frac{1}{2n} < \log \frac{2n}{n – 1} \quad \quad (1)$$ Important: this exercise is the continuation of a previous problem […]

Asymptotic behavior of $\sum\limits_{n=1}^{\infty} \frac{nx}{(n^2+x)^2}$ when $x\to\infty$

This question already has an answer here: A tricky sum to infinity 1 answer