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?

Let $a_{2n-1}=-1/\sqrt{n}$, $a_{2n}=1/\sqrt{n}+1/n$ for $n=1,2,\dots$ Show that $\prod (1+a_n)$ converges but that $\sum a_n$ diverges. What I have found so far is that $\prod_{k=2}^{2n} a_n$=$3(1-\frac{1}{2\sqrt{2}})\cdots (1-\frac{1}{n\sqrt{n}})$ and $\prod_{k=2}^{2n+1} a_n$=$3(1-\frac{1}{2\sqrt{2}})\cdots (1-\frac{1}{n\sqrt{n}})(1-\frac{1}{\sqrt{n+1}})$ I’m considering using the theorem that if each $a_n \ge 0$, then the product $\prod(1-a_n)$ converges if and only if, the series $\sum a_n$ converges. […]

$$S=\sum_{k=1}^\infty\left(\frac{k}{k+1}\right)^{k^2};\hspace{10pt}S_n=\sum_{k=1}^n\left(\frac{k}{k+1}\right)^{k^2}$$ Let $S_n$ represent its partial sums and let $S$ represent its value. Prove that $S$ is finite and find an n so large that $S_n$ approximate $S$ to three decimal places. Solution: first of all, I think that we Will use L’hopital rule and then use root test while starting to solve this. But […]

Answer to this question will greatly be appreciated if somebody solve it. Mathematica does not give the answer $$\sum_{n=1}^\infty \frac{\frac{1}{2}I_{n-1}(x)+\frac{1}{2}I_{n+1}(x)-I_{n}(x)}{n^2}$$

I was wondering if there is an explicit formulation for the series $$ \sum_{k=1}^\infty \frac{x^k}{k!\cdot k} $$ It is evident that the converges for any $x \in \mathbb{R}$. Any ideas on a formula?

Possible Duplicate: Showing that $\frac{\sqrt[n]{n!}}{n}$ $\rightarrow \frac{1}{e}$ Any suggestions to find the following limit: $$\displaystyle\lim_{n\to\infty}\left(\frac{n!}{n^n}\right)^{\frac{1}{n}}$$ with basic tools of the calculus!

It is a well known fact that: $$\lim_{N\to\infty}\sqrt[n]{n}=1$$ But what about the shifted one up: $$\lim_{N\to\infty}\sqrt[n+1]{n}=1$$ And what about the shifted one down: $$\lim_{N\to\infty}\sqrt[n]{n+1}=1$$

I.e., is there a sequence of primes whose decimal expansions have the following form: $$a_1,\ a_1a_2,\ a_1a_2a_3,\ a_1a_2a_3a_4, \dots$$ What about with the order of the digits reversed, so each number’s decimal representation is a final segment of the next one’s? (Or any other interesting variation of restrictions?) What about in other bases? In binary, […]

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