Hot answers tagged sequences-and-series
13
Hint: Note that $n^2 -(n+1)^2-(n+2)^2+(n+3)^2=4$. So by choosing signs of four consecutives appropriately, we can get $4$ or $-4$.
Any integer is of the form $4k+r$, where $r$ is one of $0$, $1$, $2$, or $3$. Now it is just a matter of showing we can get all of $1$, $2$, and $3$. And we don't even need $3$. By the way, $2$ has a length $4$ ...
12
Take logs of both sides to get
$$\log{P_n} = \sum_{i=1}^n \left(\log{\left [ 1- a\frac{i}{n} - \frac{a}{2 n}\right]} -\log{\left [ 1- a\frac{i}{n}\right]} \right) $$
Noting that $n$ is large, so we can rewrite the summand as
$$\log{\left [ 1- a\frac{i}{n} - \frac{a}{2 n}\right]} -\log{\left [ 1- a\frac{i}{n}\right]} = \log{\left( 1-\frac{a/(2 n)}{1-a ...
12
If the $2$'s were replaced by $10$'s, the problem is clear. The decimal expansion of any rational number either terminates or repeats. The number in question is one that has $1$s at all prime places after the decimal point and zeros elsewhere. This does not terminate or repeat, so it must be irrational.
The problem you have is similar, except we consider ...
7
$$\sum^\infty_{r=n}rx^{r-1}=\dfrac {d\left(\sum^\infty_{r=n}x^r\right)}{dx}$$
From here, $\sum^\infty_{r=n}x^r=\dfrac{x^n}{1-x}$ if $|x|<1$
Alternatively,
let $S_n=\sum^m_{r=n}rx^{r-1}=n\cdot x^{n-1}+(n+1)x^n+(n+2)x^{n+1}+\cdots+(m-1)x^{m-2}+mx^{m-1}$
So, multiplying by $x,$ $x\cdot S_n=\sum^\infty_{r=n}rx^r=n ...
7
Weierstrass theorem tells that on a compact interval, given a continuous function, there is a sequence of polynomials converging uniformly to this function. But the polynomials are not necessarily of the form $\sum_{j=0}^{N_n}c_jx^j$, because the coefficients depend on $n$. They are indeed of the form $\sum_{k=0}^{N_n}c_{n,k}x^k$.
To get a concrete example, ...
6
This is known as Blaschke's condition and is in fact also true for functions in the so called Nevanlinna class. The simplest way to prove this is using Jensen's formula.
Assume $f \in H^\infty$. You may as well assume that $f(0) \neq 0$ and that $f$ has infinitely many zeros. Let $n(r)$ be the number of zeros in the disc $D_r$. Fix any integer $k$ and ...
5
First, assume $f(0) \ne 0$. Let $r < 1$ and $n(r)$ be the number of zeros of $f$ inside $\overline{D}(0, r)$. Let $k$ be a positive integer so that $k < n(r)$. By Jensen's formula:
$$
\left|f(0)\right| \prod_{n=1}^{n(r)}\frac{r}{\left|\alpha_n\right|} = \exp \left\{\frac{1}{2\pi}\int_{-\pi}^\pi \log\left|f(re^{i\theta})\right|\right\}
$$
Since $f$ is ...
5
One way to show that the sum is irrational
is the well-known fact that there
are arbitrarily long intervals
where there are no primes.
If the sum were rational,
chose the interval longer that the period,
and hello contradiction!
In particular,
$n!+k$ for $k = 2$ to $n$
has no primes, since each element is divisible
by a prime from $2$ to $n$.
This also ...
5
Hint for 2)
$$\sum_{n=1}^{\infty} \frac{|\cos n|}{n} \geq \sum_{n=1}^{\infty} \frac{\cos^2 n}{n}=\sum_{n=1}^{\infty} \frac{1+\cos {2n}}{2n}$$
Convergence of $\sum_{n=1}^{\infty}\frac{\cos{2n}}{2n}$, and divergence of $\sum_{n=1}^{\infty}\frac{1}{2n}$ gives the divergence.
The same method applies to $\sum_{n=1}^{\infty}\frac{|\sin n|}{n}$.
5
Let's divide both parts by $n$. Then LHS becomes an arithmetic average of $\left(1+\frac{1}{2r}\right)^{ 2r }$, $r=1..n$. The RHS becomes (binomial coefficients) $\left(1+\frac{1}{n+1}\right)^{ n+1 }$. The function $x\to \left(1+\frac{1}{x}\right)^{ x }$ seems to be concave(need to check it, though), hence the average is not greater than the value in the ...
4
Hint:
Fix $x=-x \implies dx=-dx$
$$I=-\int_{\pi}^{-\pi}\dfrac{\sin n(-x)}{(1+(\pi)^{-x})\sin (-x)} dx$$
$$I= \int_{-\pi}^{\pi}\dfrac{\sin n(-x)}{(1+(\pi)^{-x})\sin (-x)} dx$$
$$I=\int_{-\pi}^{\pi}\dfrac{\pi^x \sin n(x)}{(\pi^x+1)\sin x} dx$$
Add this to $I=\int_{-\pi}^{\pi}\dfrac{\sin nx}{(1+\pi^{x})\sin x}dx$
$$2I= \int_{-\pi}^{\pi} \dfrac{\sin ...
4
Yes, a constant sequence (a number repeated indefinitely) is inceed monotonic: it is both monotonic non-decreasing, and monotonic non-increasing.
Hence, one can require that a sequence be strictly monotonic increasing or strictly monotonic decreasing. Under such a restriction, a constant sequence is neither strictly increasing nor strictly decreasing ...
4
\begin{align*}
\sum_{k=2}^{\infty} \frac1{k\ln^2k} &= \frac1{2\ln^22} + \frac1{3\ln^23} + \sum_{k=4}^{\infty} \frac1{k\ln^2k} \\
&\ge \frac1{2\ln^22} + \frac1{3\ln^23} + \int_4^{\infty} \frac{dx}{x\ln^2x} \\
&= \frac1{2\ln^22} + \frac1{3\ln^23} + \frac{1}{\ln4} > 2.038.
\end{align*}
(I see Andrew just wrote this in a comment.)
4
HINT:
So, the $r\ge 4$ the term $$t_r=3\cdot \frac{2^{r+1}}{5^{r-2}}=3\cdot \frac 2{5^{-2}}\cdot \left(\frac25\right)^r=150\cdot \left(\frac25\right)^r$$
Clearly, this is an infinite geometric series with the common ratio is $\frac25$ and the first term $=150\cdot \left(\frac25\right)^4$
4
$$\lim_{n \to \infty}\ln \left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^n=\lim_{n \to \infty}\frac{\ln \left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)}{\frac{1}{n}}=\lim_{x\to 0^+}\frac{\ln \left(\frac{a^{x}+b^{x}+c^{x}}{3}\right)}{x}$$
L'Hospital or the definition of the derivative solves it. Actually since the limit is just the definition of the derivative ...
4
First, the constant $c$ is slightly misleading. Why not take $d := 1-c$ so that: $$x_{n+1} = \left(1-\frac{d}{n}\right) x_{n} + \frac{b}{n}$$
Let $y_n := x_n - t$ for some $t$ that we'll establish in a second. Then:
$$ y_{n+1} + t = \left(1-\frac{d}{n}\right) y_{n} + \left(1-\frac{d}{n}\right)t + \frac{b}{n}$$
which is more naturally written as:
$$ y_{n+1} ...
4
The Cauchy Condensation Test will do it.
If you have not seen it, here is what it says. Suppose that $f(n)$ is positive nonincreasing. Then $\sum_{a}^\infty f(n)$ converges if and only if $\sum_a^\infty 2^nf(2^n)$ converges.
In our case, we have $f(n)=\frac{1}{n\log n\log\log n}$. Thus $2^n f(2^n)=\frac{1}{(n\log 2)(\log n+\log\log 2)}$. The divergence ...
3
Yes, if by non-constant you really mean that no tail of the sequence is constant, i.e., that it’s not eventually constant. If you could not do so, there would be some finite $F\subseteq\Bbb R$ such that $x_n\in F$ for each $n\in\Bbb N$. But then it must be the case that $p\in F$ and the sequence is eventually constant at $p$.
3
In fact, Goos and Norbert have given the answer. (And you should also assume $n\in \mathbb{N}$)
$$
f(n,x,y) = (x+y)^n - x^n -y^n
$$
If
both $x$ and $y$ are even: even - even - even = even;
$x$ is even and $y$ is odd: odd - even - odd = even;
$y$ is odd and $x$ is even: just like the case above.
So, $f(n,x,y)$ is always even.
3
$$u_{n+1}=\frac{1}{3}u_{n} + 4\implies 3u_n=u_n+12$$
$$\implies 3u_{n+2}=u_{n+1}+12$$
On subtraction, $$3u_{n+2}- 3u_{n+1}=u_{n+1}-u_n\implies 3u_{n+2}- 4u_{n+1}+u_n=0 $$
We can use Characteristic equation of the Recurrence relation to solve it
3
There is mre general fact. Assume $a_k\geq 0$ for all $k\in\mathbb{N}$
and $\sum_{k=1}^\infty a_k$ diverges then for all $\delta>0$ the series
$$
\sum\limits_{n=1}^\infty\frac{a_n}{S_n^{1+\delta}}\tag{1}
$$
converges. Here we denoted $S_n=\sum_{k=1}^n a_k$. Fix $n\in\mathbb{N}$ Lets apply to the function $f(t)=t^{-\delta}$ Lagrange's mean value theorem ...
3
The product can be rewritten as
$$\mathcal{P}_n=\prod_{i=1}^{n}\frac{1 - (2i + 1)a/(2n)}{1 - ia/n}=\prod_{i=1}^{n}\frac{\frac{n}{a}-\frac12-i}{\frac{n}{a}-i}=\frac{\Gamma\left(\frac{n}{a}-n\right)}{\Gamma\left(\frac{n}{a}\right)}\cdot\frac{\Gamma\left(\frac{n}{a}-\frac12\right)}{\Gamma\left(\frac{n}{a}-n-\frac12\right)}$$
Now taking the logarithm and using ...
3
The series does converge uniformly. For the proof, put $S_n(x) = \sum_{k = 0}^n \sin{(kx)}\sin{(k^2 x)}$ for $n\geq 0$. The general idea is to use summation by parts to reduce ourselves to showing that $S_n(x)$ is bounded uniformly, and then to prove that by giving a closed form for $S_n(x)$.
First, the summation by parts (I write $S_n$ in place of $S_n(x)$ ...
3
Because $ \displaystyle \lim_{n \to \infty} a_n = \infty $, it is true that $ \displaystyle\lim_{n \to \infty} \frac{a_n}{1 + a_n} = 1 $. Hence, the terms of the series do not tend to zero so it certainly does not converge.
In response to the edit, here's a brief sketch:
Assume, for the sake of contradiction, that $ \sum \frac{a_n}{1 + a_n} $ converges. ...
3
Your test of whether $a_n \gt a_{n+1}$ looks odd. Since $a_n, a_{n+1} \gt 0$, you can test whether $a_n^2\gt a_{n+1}^2$ and this comes out as $$\frac{n}{(n+4)^2}\gt \frac {n+1}{(n+5)^2}$$
It is then simple to check (clear fractions, since you are multiplying by positive numbers) that this is true for $n \ge 4$, but not for $n=3$ - which may be why the ...
3
Suppose $M$ bounds the sequence. Then, if we think of $H$ as sitting inside $H^{**}$, then for any $T \in H^\ast$ with $\|T\| \leq 1$, we have $\|x_n(T)\| = \|Tx_n\| \leq \|x_n\| \leq M$, so the operator norms of the $x_n$ thought of as operators on $H^\ast$ are bounded by $M$. Apply Banach-Alaoglu. (To the unit ball of $H^{\ast \ast}$.)
3
Some conjectural formulas for $\pi$ are given in:
J. Guillera, "Kind of proofs of Ramanujan-like series"
J. Guillera, "Mosaic supercongruences of Ramanujan-type"
J. Guillera, "About a new kind of Ramanujan type series"
For example (I expressed an infinite sum given in the paper in terms of hypergeometric functions):
$$224 \,
...
3
The method of residues applies to sums of the form
$$\sum_{n=-\infty}^{\infty} f(n) = -\sum_k \text{res}_{z=z_k} \pi \cot{\pi z}\, f(z)$$
where $z_k$ are poles of $f$ that are not integers. So when $f$ is even in $n$, you may express as follows:
$$2 \sum_{n=1}^{\infty} f(n) + f(0)$$
For this case, $f(z)=1/(z^2+a^2)$ and the poles $z_{\pm}=\pm i a$ and ...
3
Set $a_k = (k-1)!!/k!!$, so that we're looking at the series $\sum_{k=1}^\infty (-1)^k a_k$. Notice that the odd terms are exactly the normalized central binomial coefficients:
$$
a_{2n} = \frac1{4^n} \binom{2n}n.
$$
By known asymptotics, $a_{2n} \sim 1/\sqrt{\pi n}$ as $n$ grows large. Similarly,
$$
a_{2n-1} = \frac1{2na_{2n}} \sim \frac{\sqrt{\pi n}}{2n} = ...
3
In a sense, the proof writes itself. The details depend on whether we use $\lt$ or $\le$ in the definition of monotonically decreasing. Let's use $\le$. We want to know whether
$$\frac{a_1+a_2 +\cdots+a_{n+1}}{n+1} \overset{?}{\le} \frac{a_1+a_2+\cdots+a_n}{n}.\tag{$1$}$$
Algebraic manipulation shows that this is equivalent to
$$na_{n+1}\overset{?}{\le} ...
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