New answers tagged sequences-and-series
1
The assertion
$$ \lim_{x\to\infty} \sin\frac{2\pi}{3^{n}} \sim \frac{2\pi}{3^{n}}$$
is perhaps not precise enough for our needs. But it can be made precise, in a form useful for Ratio Test calculations.
We have
$$\lim_{t\to 0}\frac{\sin t}{t}=1.$$
We can replace $\sin\left(\frac{2\pi}{3^n}\right)$ by
...
1
$\lim_{n\to\infty} \sin\frac{2\pi}{3^{n}} \sim \frac{2\pi}{3^{n}}\tag{3}$ Is correct since $\lim_{x\to 0 } \dfrac{sinx}{x}=1$ and since sinus is continuous this holds for any sequence
${x_n\to 0}$
1
$$1+(\frac12+\frac12)+ (\frac14+\frac14+\frac14+\frac14)+ \dots (\dots \frac1{2^{n-1}})\gt$$
$$\frac{1}{1} + \frac{1}{2} + \frac{1}{3}+ \frac{1}{4}+...+ \frac{1}{2^n-1}\gt$$ $$1+\frac12+(\frac14+\frac14)+(\frac18+\frac18+\frac18+\frac18)+\dots +( \dots \frac1{2^n})$$ (compare equivalent terms, and count the elements in the last bracket carefully) gives ...
0
$f(n)=H_{2^n-1}$, the $(2^n-1)$-st harmonic number. There is no closed form, but link gives the excellent approximation
$$H_n\approx\ln n+\gamma+\frac1{2n}+\sum_{k\ge 1}\frac{B_{2k}}{2kn^{2k}}=\ln n+\gamma+\frac1{2n}-\frac1{12n^2}+\frac1{120n^4}-\ldots\;,$$
where $\gamma\approx 0.5772156649$ is the Euler–Mascheroni constant, and the $B_{2k}$ are Bernoulli ...
2
$$f(n)\approx \ln(2^n-1)+\gamma\approx n\ln 2+\gamma$$
-1
$$\large \lim_{n \to \infty}\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^{n}=\ \lim_{r \to 0}\left(\frac{a^{r}+b^{r}+c^{r}}{3}\right)^{1/r}\\ =\huge \lim_{r \to 0}\frac{1}{r}\cdot e^{[log \left(a^{r}+b^{r}+c^{r}\right)]}$$
$$=\huge e^{[\lim_{r \to 0}\frac{d}{dr}log(a^{r}+b^{r}+c^{r})]}$$
$$=\huge e^{\frac{[log(a)+log(b)+log(c)]}{3}= (abc)^{1/3}}$$
3
$$\lim_{n \to \infty}\ln \left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^n=\lim_{n \to \infty}\frac{\ln \left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)}{\frac{1}{n}}=\lim_{x\to 0^+}\frac{\ln \left(\frac{a^{x}+b^{x}+c^{x}}{3}\right)}{x}$$
L'Hospital or the definition of the derivative solves it. Actually since the limit is just the definition of the derivative ...
1
You know that $x_n\to 3/2$ as $n\to\infty$. You are doing fixed point (Picard) iteration for the function
$$f(x) = \sqrt{(x + 3)\over 2}$$
converging to $x = 3/2$ and you know that $$f'(3/2) = {1\over 6\sqrt{2}}.$$
Does this tell you anything?
2
My guess would be they want something like this (as examples): the "turning pattern"
$$\underbrace{\fbox{$\,\uparrow\,\strut$}\;\fbox{$\rightarrow\strut$}\;\fbox{$\,\downarrow\,\strut$}\;\fbox{$\rightarrow\strut$}}_{\large\mathtt{part\, that\, ...
0
Let there is an A.P. whose first term is a and common diff. is d.
$$S_q=\frac q2\left(2a+(q-1)d\right)$$
$$p=\frac q2\left(2a+(q-1)d\right)\cdots(1)$$
$$S_p=\frac p2\left(2a+(p-1)d\right)$$
$$q=\frac p2\left(2a+(p-1)d\right)\cdots (2)$$
solve (1) and (2)
$$d=\dfrac{-2(p+q)}{pq}$$
thus
$$S_{p+q}=\frac{p+q}2\left(2a+(p+q-1)d\right)$$
$$S_{p+q}=\frac ...
0
HINT
The sum of first $n$ terms of AP with first term $a_1$ and difference $d$ is
$$S_n=\frac{n}{2}(2a_1+(n-1)d)$$
in case we need to solve the system
$$p=S_q=\frac{q}{2}(2a_1+(q-1)d)$$
$$q=S_p=\frac{p}{2}(2a_1+(p-1)d)$$
you need to expres $a_1$ and $d$ as functions of $p$ and $q$
0
Let the A.P. be $\,a_1,a_2,...\;,\;\;d:=\text{the constant difference}\,$ , then
$$S_q=\frac q2\left(2a_1+(q-1)d\right)=p$$
$$S_p=\frac p2\left(2a_1+(p-1)d\right)=q$$
thus
$$S_{p+q}=\frac{p+q}2\left(2a_1+(p+q-1)d\right)=\frac p2\left(2a_1+(p-1)d\right)+\frac{pqd}2+\frac q2\left(2a_1+(q-1)d\right)+\frac{pqd}2$$
...and etc.
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 $H^{\ast \ast}$.)
0
(This is a corrected version)
I get that the sum converges,
but do not know the value.
Since $!n = n!\sum_{k=0}^n\frac{(-1)^k}{k!}$,
$\frac{!n \cdot e}{n!}
= e \sum_{k=0}^n\frac{(-1)^k}{k!}
= e \big(\frac1{e}-\sum_{k=n+1}^{\infty}\frac{(-1)^k}{k!}\big)
= 1-e\sum_{k=n+1}^{\infty}\frac{(-1)^k}{k!}
$.
Starting as Carl Najafi did,
$\begin{align*}
...
4
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 ...
0
Some calculations: We have
$$
\sum_{n=2}^\infty\left|\frac{!n\cdot e}{n!}-1\right|=\sum_{n=2}^\infty\left|\sum_{k=0}^\infty e\frac{(-1)^k}{k!}-1\right|=\sum_{n=2}^\infty\left|\frac{\Gamma(n+1,-1)}{\Gamma(n+1)}-1\right|.
$$
We also have
$$
...
12
If the $2$'s were replaced by $10$'s, the problem is clear. The decimal expansion of any rational number either terminates or repeats. The number in question is one that has $1$s at all prime places after the decimal point and zeros elsewhere. This does not terminate or repeat, so it must be irrational.
The problem you have is similar, except we consider ...
17
Hint: Consider this number in base 2. If it is rational, it must have a period. What sort of consequence does this mean, that would lead to a contradiction?
2
There is another missing piece (in response to the comments) that must be added here, and that is the proof of the claim that $$\frac{1}{\pi^s}\Gamma(s)(s-1/4)\zeta(2s)$$ is odd and purely imaginary on the line $\Re(s) = 1/4.$ This is due to a deep connection to the functional equation of the Riemman zeta function. We will show that ...
3
The method of residues applies to sums of the form
$$\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)$$
For this case, $f(z)=1/(z^2+a^2)$ and the poles $z_{\pm}=\pm i a$ and ...
2
Essentially, you want to "round" $a_n$ down to the greatest number which can be expressed as $a_0 + kx$ for some integer $x$. One way to do so is to replace $a_n$ by $(a_0+x\lfloor\frac{a_n-a_0}{x}\rfloor)$ in your summation formula.
4
\begin{align*}
\sum_{k=2}^{\infty} \frac1{k\ln^2k} &= \frac1{2\ln^22} + \frac1{3\ln^23} + \sum_{k=4}^{\infty} \frac1{k\ln^2k} \\
&\ge \frac1{2\ln^22} + \frac1{3\ln^23} + \int_4^{\infty} \frac{dx}{x\ln^2x} \\
&= \frac1{2\ln^22} + \frac1{3\ln^23} + \frac{1}{\ln4} > 2.038.
\end{align*}
(I see Andrew just wrote this in a comment.)
3
Hint :
Prove that the sequence $\{x_n\}$ is bounded and monotonic
1
(Of couse we need $a\ge 0$ as premise)
Show by induction that $0\le x_n<a+1$
Show by induction that $x_{n+1}\ge x_n$
(You can facilitate things a bit by prepending a term $x_0=0$. Not that then still $x_1=\sqrt{a+x_0}$)
1
First, this sum is an instance of the binomial theorem and $\binom{k}{k-1}=(-1)^{k-1}\binom{-2}{k-1}$
$$
\begin{align}
\sum_{k=1}^\infty kx^k
&=\sum_{k=1}^\infty\binom{k}{1}x^k\\
&=\sum_{k=1}^\infty\binom{k}{k-1}x^k\\
&=\sum_{k=1}^\infty(-1)^{k-1}\binom{-2}{k-1}x^k\\
&=\sum_{k=0}^\infty(-1)^k\binom{-2}{k}x^{k+1}\\
...
2
It works like @vadium123 mentioned, I just wanna give you the idea how one derives that result.
The first idea is that
\[ \sum_{n=0}^\infty n \cdot x^n \]
looks like the goemetric series (at least a bit). In fact if we take the derivative the geometric series in respect to $x$ we get
\[ \sum_{n=1}^\infty n\cdot x^{n-1} \]
So when we take
\[ x \cdot ...
2
We have, provided $|a|<1$, that $$\sum_{n=0}^\infty na^n=\frac{a}{(a-1)^2}$$
In your problem, we may rewrite $$\sum_{x=1}^\infty p(1-p)^x(2xy-x)=(2y-1)p\sum_{x=1}^\infty x(1-p)^x$$
Applying that first identity, we find the sum to be $$\frac{(2y-1)p(1-p)}{p^2}=\frac{(2y-1)(1-p)}{p}$$
Edit: Deriving the formula, per request.
We start with the geometric ...
2
It is a fact of life that homographic transforms like
$$
u:x\mapsto R+\frac{Rx}{R+x},
$$
are conjugate to affine transforms. The first step to make use of this fact for the iterates of $u$ is to determine a fixed point of $u$, here
$$
x^*=R\frac{1+\sqrt5}2,
$$
and to rewrite everything in terms of $1/(x-x^*)$, here
$$
\frac1{u(x)-x^*}=b+\frac{a}{x-x^*},
$$
...
1
Let $R_n$ be the resistance of a circuit with $2n$ resistors, so $R_1 = 2R$. The circuit with $2(n+1)$ resistors is obtained from the circuit with $2n$ resistors by placing one resistor in parallel and another one in series. So
$$
R_{n+1} = R + \frac{1}{\frac{1}{R} + \frac{1}{R_n}} = R + \frac{R R_n}{R+R_n}.
$$
Substituting $R_n = \frac{p_n}{q_n}R$ gives
...
1
The second question is easier. Suppose that the resistance for infinite number of elementary parts is equal to $R_{\infty}$. Then it should not change if you add one more elementary part. This gives the equation
$$R_{\infty}=\frac{RR_{\infty}}{R+R_{\infty}}+R,$$
with the solution $\displaystyle R_{\infty}=\frac{1+\sqrt{5}}{2}R$.
Now concerning the first ...
1
Just notice that,
if $\sum_{n\geq 1}|a_n|<\infty$ , then $\sum_{n\geq 1}a_n<\infty$. Also, if $\sum_{n\geq 1}a_n$ converges, then $\sum_{n\geq 1}\frac{a_n}{n}$ converges too by Dirichlet's test, so we get
$$ \sum_{n \geq1}^{\infty}\frac{n+1}{n}a_n = \sum_{n \geq 1}a_n + \sum_{n \geq 1}\frac{a_n}{n}< \infty.$$
4
Hints:
Show that, in fact, $\sum\limits_{n=1}^\infty {n+1\over n} a_n$ converges absolutely. This will imply the series converges.
To show absolute convergence, you may use the Comparision test. Note that for each $n$, one has ${n+1\over n}=1+{1\over n}\le 2$.
2
The sequence increases in wide sense means that it is not strictly increasing ( $a_n>a_{n-1})$ but it is non-decreasing $a_n\geq a_{n-1}$
2
think about $f(x) = \frac 1x$.
On any closed interval $[a,b]$ with $a>0$ the function is bounded and I can approximate it very well with a polynomial.
But on the open interval $(0,1)$ it's unbounded. The reason I can get away with this is because it's not defined at $0$. Any continuous function on a closed interval is bounded. That's not true on ...
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, ...
2
We want to show that: if the series $\sum_{n=1}^{\infty}a_{n}^2$ diverges then the $\sum_{n=1}^{\infty}|a_{n}|$ diverges.
To prove this use that $a^2_n<1$ for all $n>N$ for some $N\in\mathbb N$ and therefore $|a_n|>a_n^2$ for all $n>N$...
1
And here's the proof why $(1+1/x)^x$ is concave for $x\le 1$ (the case $x\ge 1$ is already proven by @TCL).
We study the second derivative with respect to $x$, it is equal to (thanks to TCL) to
$\displaystyle \left(1+ \frac 1x\right)^x\left(\left(\ln (1+1/x)-\frac{1}{1+x}\right)^2-\frac{1}{x(1+x)^2}\right)$;
First of all, $\ln (1+1/x)\ge\frac{1}{1+x}$; ...
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 $\sum_{n=1}^{\infty}\frac{|\sin n|}{n}$.
2
Here is the proof that $(1+1/x)^x$ is concave for $x\ge 1$.
The second derivative of $(1+1/x)^x$ is $(1+1/x)^x$ times $$p(x)=\left(\ln\left(1+\frac{1}{x}\right)-\frac{1}{1+x}\right)^2-\frac{1}{x(1+x)^2}$$
Now for $x\ge 1$, we have $$\ln(1+1/x)-\frac{2}{1+x}\le \frac{1}{x}-\frac{2}{1+x}=\frac{1-x}{x(1+x)}\le 0$$ and $$\ln(1+1/x)\ge ...
5
Let's divide both parts by $n$. Then LHS becomes an arithmetic average of $\left(1+\frac{1}{2r}\right)^{ 2r }$, $r=1..n$. The RHS becomes (binomial coefficients) $\left(1+\frac{1}{n+1}\right)^{ n+1 }$. The function $x\to \left(1+\frac{1}{x}\right)^{ x }$ seems to be concave(need to check it, though), hence the average is not greater than the value in the ...
2
The geometric series $\sum_{k\geq k_0} r^k$ converges if and only if $|r|<1$. You have a confusion regarding its actual sum, which depends where the summation starts. In particular
$$
\sum_{k\geq 1}r^k=\frac{r}{1-r}\qquad\mbox{and}\qquad\sum_{k\geq 0}r^k=\frac{1}{1-r}.
$$
More generally,
$$
\sum_{k\geq k_0}r^k=\frac{r^{k_0}}{1-r}.
$$
To see where this ...
0
There are two conditions for an alternating series to be convergent.
1) For some particular n, the corresponding graph must be declining and by using the derivative, this is easy to verify.
2) The infinite term must go towards zero, which it does.
If you remove the (-1) part, the series will diverge, hence your series is Conditionally Convergent. ...
2
No. Hint: The right-hand-side $\int_0^1 x^{n+k} dx=c_k$ is a constant for each $k$. You can make $f_k(x)$ completely arbitrary on the interval $[0,9/10]$, and then adjust it on the interval $[9/10,1]$ to give the integral $\int_0^1 f_k(x)x^n dx$ the value $c_k$. Choose $f_k$ on the interval $[0,9/10]$ in such a way that the series diverges for any $0\leq ...
2
I usually use "initial segment", and "tail segment" or "end segment" or "final segment" to denote the last part of a sequence.
If you want to indicate the segment is not everything, then "proper initial segment" should suffice.
0
In computer science (which is, at least at the beginning, maths), you call them words or strings over an alphabet $X$ which contains your letters.
For this alphabet $X$, you define $X^n$ by the usual Cartesian product and $X^*=\bigcup\limits_{n\in \Bbb N}X^n$. $X^*$ is the set of all words written with letters of $X$.
Then you can define a product $\cdot$ ...
1
If you are only interested in finding the limit, then here is an algebraic way to find it. Assume
$$\lim_{n\to \infty} u_n = a \implies \lim_{n\to \infty} u_{n+1} = \lim _{n\to \infty }(1+\frac{1}{u_n}) \implies a = 1+\frac{1}{a}\implies a^2-a-1=0 $$
$$ \implies a = \dots\,. $$
Note that, the terms of the sequence are positive.
2
Writing $u_n$ as $\frac{p_n}{q_n}$ for two sequences $(p_n)$, $(q_n)$ to be determined. We have:
$$u_{n+1} = 1 + \frac{1}{u_n} \iff \frac{p_{n+1}}{q_{n+1}} = 1 + \frac{q_n}{p_n} = \frac{p_n+q_n}{p_n}$$
Normalize $(p_n)$ and $(q_n)$ appropriately, we can turn the non-linear recurrence equation into a linear one:
...
2
If you want to be rigorous you should proove that the limit exists.
You can do that showing that the odd terms form a decreasing sequence while the even terms form an increasing sequence (induction). Then show that those sequences are bounded below/above.
1
I just found out how to do it. Assume that the limit is equal to a. Therefore, the limit of u_n+1 and u_n are both equal to a. Using the recursive definition, place a back in to get. a=1+1/a which is equivalent to a^2-a-1=0. Solving for a gives two answers. But, since u_n is bounded between 1 and 2; (1+sqrt(5))/2 is the accepted limit.
5
The simplest example I can think of is $\{1,0,1,0,...\}$. If you want your elements to be strictly positive, use some fast-converging sequence such as $n^{-4}$ in place of the zeroes.
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