I am having trouble finding the correct binomial expansion for $\dfrac{1}{\sqrt{1-4x}}$: Simplifying the radical I get: $(1-4x)^{-\frac{1}{2}}$ Now I want to find ${n\choose k} = {\frac{-1}{2}\choose k}$ \begin{align} {\frac{-1}{2}\choose k} &= \dfrac{\frac{-1}{2}(\frac{-1}{2}-1)(\frac{-1}{2}-2)\ldots(\frac{-1}{2}-k+1)}{k!} \\ &= (-1)^k\dfrac{\frac{1}{2}(\frac{1}{2}+1)(\frac{1}{2}+2)\ldots(\frac{1}{2}+k-1)}{k!} \\ &=(-1)^k{\frac{1}{2}+k-1\choose k} \end{align} Now, applying Newton’s Generalized Binomial Theorem I get: \begin{equation}\sum\limits_{k=0}^\infty(-1)^k{\frac{1}{2}+k-1\choose k}(4x)^k \end{equation} Is this correct? it seems […]

For a sequence $a_n = O(n^{-1/2})$ as $n\to\infty$, consider the corresponding Cesàro means $b_n = \frac{1}{n} \sum_{j=1}^n a_j$. Is it possible to derive the rate of convergence for the sequence $b_n$? What about the general case $a_n = O(c_n)$?

This question already has an answer here: What's the value of $\sum\limits_{k=1}^{\infty}\frac{k^2}{k!}$? 5 answers

This problem is from calculus-2 course. The basic knowledge includes integral test and $p$-series test. Find an $N$ so that $$\sum_{n=1}^\infty {1\over n^4}$$ is between $$\sum_{n=1}^N {1\over n^4}$$ and $$\sum_{n=1}^N {1\over n^4} + 0.005$$ The Answer is $N=5$. How to solve it? Thanks!

T. Gunn’s answer to a question about linear combinations being restricted to finite sums asserted that For example, you would not count the “infinite” linear combination $$\exp(x) = \sum_{n = 0}^\infty \frac{x^n}{n!} $$ to be in the span of $\{1,x,x^2,\dots\}$ in the vector space $C[0,1]$ of continuous functions on $[0, 1]$. After some brief reading, […]

Riemann’s theorem states that if a series is conditionally convergent, then for any number $L$ (could be infinite), the series can be rearranged in such manner that it would converge to $L$. I was wondering, is the converse true? More formally, let $a_n$ be a sequence such that $a_n\to 0$, and for every $L$ (could […]

assume {$a_n$}$_{n=1}^\infty$ ,$a_n$ is none negative and real sequence that satisfied :$$1+a_{m+n}\leq (1+a_{m})(1+a_{n}) ,\quad m,n\in\mathbb N$$ how prove $x_n=(1+a_{n})^\frac1n $ is convergent? thanks in advance

If the first term in our geometric progression (GP) is $k$, and the common ratio is 0, then our sequence is $\{k, 0, 0, 0, 0,\ldots\}$. Is there anything wrong with this statement? So, is $\{0, 0, 0,\ldots\}$ a GP? I have googled for a definition of GP, but wikipedia (which I am skeptical about) […]

In the book “General topology” of Lipschutz, exercise 26 at page 222 says that: if $(f_n)$ be a sequence of real value differentiable funtions on $[a, b]$ which converge uniformly to $g$ then $$\lim_{n\to \infty} \frac{d}{dx}f_n(x)= \frac{d}{dx}\lim_{n\to \infty}f_n(x).$$ But I find that the sequence $$f_n(x)=\frac{1}{n}\cos(nx), \ \ x\in [0, 1]$$ satisfying $f_n$ converge uniformly to […]

So I got this challenge from my teacher. Solve ${x^{x^{x^{x^{x^{\dots}}}}}} = y$ (eq. 1) for $x$. My attempt: As $x^{y^z}$ per definition equals $x^{y \cdot z}$, then $x^y = y$ from (eq. 1). Thus, $x = \sqrt{y}$. Is this a valid proof?

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