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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... 2S:=\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}+\cdotsS-\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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