# Tag Info

## Hot answers tagged sequences-and-series

4

You're on the right track, but you gave up too much in the numerator. Instead, observe that $$\frac{2\sqrt{n}+1}{n^2+n+1}\leq \frac{2\sqrt{n}+1}{n^2}\leq \frac{3\sqrt{n}}{n^2}=\frac{3}{n^{\frac{3}{2}}}$$ and $\sum_{n=1}^{\infty}\frac{3}{n^{\frac{3}{2}}}$ converges.

4

The modified Bessel functions of the first kind fulfill the recurrence relation: $$I_n(\alpha)=\frac{\alpha}{2n}\left(I_{n-1}(\alpha)-I_{n+1}(\alpha)\right)\tag{1}$$ due to the integral representation: $$I_n(\alpha) = \frac{1}{\pi}\int_{0}^{\pi}\cos(nx)\,e^{\alpha\cos x}\,dx \tag{2}$$ and the cosine addition formulas. A straightforward consequence of $(1)$ ...

3

Let $x_n = {1 \over n} \lfloor nx \rfloor$. Then $x_n$ is rational for all $n$ and $x_n \to x$ for any $x$. There is nothing special about the sequence $n$, any rational sequence $q_n$ such that $q_n \to \infty$ will do, that is $x_n = {1 \over q_n} \lfloor q_n x \rfloor$.

2

Identify $\Bbb R^2$ with $\Bbb C$. Then $(\cos x,\sin x)$ becomes $\cos x+i\sin x=e^{ix}$ and so $$g(\alpha)=\sum_{n=1}^\infty {(-1)^{n+1}}n^{-\alpha+it}$$ Maybe that halps.

2

The value of $p$ is different for each of the inequalities. The inequality $\displaystyle \int_1^{n+1} \frac{1}{x}\, dx \le \sum_{k=1}^n \frac{1}{k}$ comes from taking $p=0$ and $q=n$ in the theorem. The inequality $\displaystyle \sum_{k=2}^n \frac{1}{k} \le \int_1^n \frac{1}{x}, dx$ comes from taking $p=1$ and $q=n$ in the proposition. (It should also be ...

2

The reason for the $n+1$ in $\displaystyle \int_1^{n+1} \frac{1}{x}\, dx \le \sum_{k=1}^n \frac{1}{k}$ can be seen from this diagram from Wikipedia so $\displaystyle \int_1^{6} \frac{1}{x}\, dx \le \frac11+\frac12+\frac13+\frac14+\frac15$ In effect they took $p=0, \, q=n$ for the left hand inequality, possibly because it is more natural, and then they ...

2

Suppose $\{a_n\}$ is the sequence of terms of this series. Then it is clear that $\lim \sup (|a_n|)^{1/n} = 1$ so that the radius of convergence is $1$.

2

Taking the absolute value of your algebra gives $x^2$. Since the ratio test looks for when this is $<1$, we ask ourselves when is $|x^2|<1$. The answer is when $-1<x<1$, and if $|x|>1$ it does not converge. This suffices to find the radius; it is $1$. If you want the interval of convergence as well, then you need to test when $|x|=1$. ...

1

$P'(x)=1-x^2+x^4-x^6+\ldots$ is a geometric series, and converges to $\frac{1}{1+x^2}$ for $|x|<1$.

1

For each particular $b$, it will have a closed form: since the $\sin$ term depends only on $n \mod (2b)$, this reduces to the sum of $2b$ geometric series. I doubt that there is a closed form expression as a function of $a$ and $b$.

1

The function $$f(x)=\frac{1}{x}$$ Is convex on $(0,\infty)$, so for $a_1,\ldots,a_n>0$ we have $$\frac{n}{a_1+\cdots+a_n}\leq\frac{1}{na_1}+\cdots+\frac{1}{na_n}.$$ Rearranging, we obtain the result.

1

You have already found that: $\forall \ n, a_n+b_n+c_n=1$ Without loss of generality, suppose that: $a_{n}\leq b_{n}\leq c_{n}$. Then: $a_{n+1}=a_{n}^{2}+2b_{n}c_{n}=a_{n}^{2}+b_{n}c_{n}+b_{n}c_{n}\leq a_{n}c_{n}+b_{n}c_{n}+c_{n}c_{n}=c_{n}(a_{n}+b_{n}+c_{n})=c_{n}$. Similary, b_{n+1}=b_{n}^{2}+2a_{n}c_{n}\leq b_{n}c_{n}+a_{n}c_{n}+c_{n}^{2}=c_{n}(a_{n}+... 1 You do not really need the integral test to prove it. Now that Landau's notation is your forte, you may simpy notice that $$\text{arctanh}\frac{1}{k}=\frac{1}{2}\log\left(\frac{1+\frac{1}{k}}{1-\frac{1}{k}}\right)=\sum_{n\geq 0}\frac{1}{(2n+1)\,k^{2n+1}}=\frac{1}{k}+O\left(\frac{1}{k^3}\right)\tag{1}$$ implies: $$H_n = \sum_{k=1}^{n}\frac{1}{k} = O(1)+\... 1 For conditionally convergence use the Leibniz convergence theorem, the sequence is decreasing and converges to zero. But the series does not converge absolute:$$\frac{\sqrt{k}}{k+1}\geq \frac{1}{k+1}which does diverge. 1 The syntax is mostly correct. The only improvement I see is to write the k-th derivative as f^{(k)}, not f^k. This also applies when k is a known number. There is an alternative notation with apostrophes. So you should write f^{(0)}(x) or f(x), f^{(1)}(x) or f'(x), f^{(2)}(x) or f''(x) and so forth. The second one is generally only used ... 1 We can use the standard formula with a slight variation: We obtain \begin{align*} \left(\sum_{k=0}^\infty a_kx^{2k}\right)\left(\sum_{l=0}^\infty b_lx^l\right) &=\sum_{n=0}^\infty\left(\sum_{{2k+l=n}\atop{k,l\geq 0}}a_kb_l\right)x^n\tag{1}\\ &=\sum_{n=0}^\infty\left(\sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor}a_kb_{n-2k}\right)x^n\tag{2} \... 1 This particular one is I_1(2)/I_0(2) where I_0 and I_1 are modified Bessel functions of the first kind, of orders 0 and 1. 1 Yes, you should add \Delta t \to 0 to be explicit, but in general, o(\Delta t) is understood to hold for \Delta t \to 0, so I think it is fine not to add it, but there is no harm in writing it. To both notations you give: They talk about different things. The first one is the definition of y being differentiable at t, iff y(t + \Delta t) = y(... 1 This is partial fraction decomposition. If you have $$\frac x {(x-a)(x-b)}=\frac A {x-a}+\frac B {x-b}$$ that is to say $$x=A(x-b)+B(x-a)=(A+B) x-(Ab+Ba)$$ Comparing terms, you then have equations $$1=A+B$$ $$0=Ab+Ba$$ Solve them forA,B\$ to get $$\frac x {(x-a)(x-b)}=\frac{a}{(a-b) (x-a)}-\frac{b}{(a-b) (x-b)}$$

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