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5

Your sequence converges to $$J_1(1) = \frac{1}{2\pi} \int_0^{2\pi} \sin(t+\cos t) \, \mathrm{d} t \approx 0.440051. $$ The series expansion of the Bessel function gives an alternative expression for this constant: $$J_1(1) = \sum_{m=0}^\infty \frac{(-1)^m}{m!(m+1)!2^{2m+1}}.$$ This expression easily implies that $J_1(1)$ is irrational. Consider first the ...


5

If $|b_n| \leq M$ for all $n$, then $|a_nb_n| \leq |a_n|M$ for all $n$.


3

By the Taylor series we have $$\left(1+\frac1n\right)^n=\exp\left(1-\frac1{2n}+\mathcal O\left(\frac1{n^2}\right)\right)=e-\frac e{2n}+\mathcal O\left(\frac1{n^2}\right)$$ so we see that this series is convergent since $\sum\frac{(-1)^n}{n}$ is convergent by Leibniz theorem and the series $\sum O\left(\frac1{n^2}\right)$ is convergent by comparison with the ...


3

This only works if the series converges. If $\|A\|<1$ and $\|\cdot\|$ is a submultiplicative norm, then the result follows from a geometric series approach. We have $\| \sum_{k=0}^n A^k \| \le \sum_{k=0}^n \|A\|^k \le {1 \over 1 - \|a\|}$, hence the series converges. We have $(I-A) \sum_{k=0}^n A^k = I-A^{n+1}$, hence letting $n \to \infty$, we have ...


2

I think your observation helps if you iterate it more. $$ \begin{align} S_m&=2-m\sum_{n=0}^\infty\frac{n}{2^n(2^n+m)}\\ &=2-m\sum_{n=0}^\infty \frac{n}{2^n}\left(\frac{1}{2^n}-\frac{m}{2^n(2^n+m)}\right)\\ &=2-m\sum_{n=0}^\infty \frac{n}{4^n}+m^2\sum_{n=0}^\infty\frac{n}{4^n(2^n+m)}\\ &=2-\frac49m+m^2\sum_{n=0}^\infty\frac{n}{4^n(2^n+m)}\\ ...


2

Hint: if $x_n\to x$ then $x_{n+1}-x_n \to 0$. But...


2

$$S:=\sum_{n=1}^\infty\frac{(H_n)^3}{n^3}$$ is convergent since the terms are equivalent to $\,\left(\dfrac{\ln(n)}n\right)^3\,$ as $\;n\to \infty\,$ and given by $$S=\frac {31}{5040}\pi^6-\frac 52\zeta(3)^2\\\approx 2.3009545517005250398$$ I don't know a complete proof but we may start with the following class of Euler sums : ...


1

$\exists{\varepsilon}>0, \forall{N∈\Bbb{N}}, \exists{n}>N : |x_n-x|\ge\varepsilon$ Let's take $\varepsilon = \frac{1}{2}$. $\forall n > 1: |x_n| \geq \frac{1}{2}$ and $|x_n - x_{n+1}| \geq 1$. Then even if for some $n$: $|x_n - x| < \varepsilon$ then $|x_{n+1} - x| = |x_{n+1} - x_n + x_n - x| \geq |x_{n+1} - x_n| - |x_n - x| > 1- ...


1

Hints: Squeeze Rule and $$ | \sin \alpha | \leq 1 $$ for all $\alpha $


1

If $norm(A)<1$ then the series $C=1+A+A^2+A^3+...$ converges to $C=\frac{1}{(I-A)}$ which is the definition of inverse of $C$ i.e. $C=(I-A)^{-1}$ It is kind of like geometric series.


1

Hint : How about noticing this pattern $$S=\sum ^{\infty}_{n=1} {e}^{-n}=e^{-1}+e^{-2}+e^{-3}+\cdots$$ $$\frac{S}{e}=e^{-2}+e^{-3}+e^{-4}+\cdots$$ $$S-\frac{S}{e}=e^{-1}$$ $$S\left(1-\frac{1}{e}\right)=\frac{1}{e}$$ $$S=\frac{1}{e-1}$$


1

$$\frac{3pp_n^2-2p_n^3-4}{3pp_n^2-3p_n^3}=\frac{3pp_n^2-2p_n^3-p^3}{3pp_n^2-3p_n^3}\\ =\frac{(p_n-p)(p^2+pp_n-2p_n^2)}{3p_n^2(p-p_n)}$$ I think you needed $E_{n+1}/E_n^2$ because Newton's method tends to double the number of digits precision.


1

Hint: you cannot write as telescopic, since you do not have hypothesis on $a,b$. $$\frac{1}{n+a-b}-\frac1{n+a}=\frac{b}{(n+a-b)(n+a)}, $$ hence $$\sum \Big(\frac{1}{n+a-b}-\frac1{n+a}\Big)=\sum\frac{b}{(n+a-b)(n+a)},$$ which converges since is comparable with $\sum\frac1{n^2}$...


1

Hint What is $AC$? What happens if you subtract $C-AC$? Note that $C-AC=(I-A)C$.


1

Converge absolutely? WRONG. Because as you said, $e-\left(1+\frac{1}{n}\right)^n > \frac{1}{2n}$. Let us prove this inequality. \begin{align*}e-\left(1+\frac{1}{n}\right)^n>{}&\left(1+\frac{1}{2n}\right)^{2n}-\left(1+\frac{1}{n}\right)^n \\ ={}&\left(1+\frac{1}{n}+\frac{1}{4n^2}\right)^{n}-\left(1+\frac{1}{n}\right)^n\\ ...


1

It is okay. However, you should know that $$\sum_{n=-\infty}^{+\infty} f(x,n)=\lim_{n\to +\infty} \sum_{k=-n}^{n}f(x,k)$$


1

On second thought, this is not weirder than writing $+\infty$ on the top. Recall that the notation $\sum_{n=n_0}^\infty$ is not a sum but a series, i.e. already there we have a notational distinction from $\sum_{n=n_0}^{n_1}$ where asummand with $n=n_1$ does occur in the sum, whereas no summand with $n=\infty$ "occurs" in the series $\sum_{n=n_0}^\infty$. A ...



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