New answers tagged

1

The key aspect is to generalize this for all $n \in \mathbb{N} \cup \{0\}.$ If we wanted to prove that for some arbitrary $n \in \mathbb{N} \cup \{0\},$ $$s_{n+1} > s_n$$ $$\leftarrow \sum_{k=0}^{n+1} \frac{1}{k!} > \sum_{k=0}^{n} \frac{1}{k!}$$ $$\leftarrow \frac{1}{(n+1)!} + \sum_{k=0}^n \frac{1}{k!} > \sum_{k=0}^n \frac{1}{k!}$$ $$\leftarrow ...


1

You could prove this by induction. First, prove the base case: the summation when $n=2$ is greater than the summation when $n=1$ Then, assume this is true for all $m<n$. Need to show this holds for $n+1$ and $n$. $\sum_{k=0}^{n+1}\frac{1}{k!}=(\sum_{k=0}^{n}\frac{1}{k!})+\frac{1}{(1+n)!}>\sum_{k=0}^{n}\frac{1}{k!}$ So it holds for $n+1$ and $n$. ...


2

You know that at $25$m the temperature is $10°$, so you need to gain $20°$ to reach $30°$, since you gain $1°$ by $32$m, to gain $20°$ you need to add $32 \times 20=640$m. So finally you reach $30°$ at $640+25=665$m. For the second question, you are $960$m deeper that $25$m, since $960=32 \times 30$, you gained $30°$ relatively to the temperature at $25$m ...


2

An arithmetic series is a synonym for a linear equation, only the naming is more complicated. The slope of the function $f(x)=ax+b$ must be $\frac{1}{32}$ because $32m$ rise the temperature by $1$ degree. So, we have $f(x)=\frac{1}{32}x+b$. We also know $f(25)=10$, if we insert this, we get $b=\frac{295}{32}$ So, the temperature is $y=\frac{x+295}{32}$ ...


0

Let $O$ be the set of observations. Let $T$ be the set of timestamps ordered by a partial order $\leq$. We associate $O$ to $T$ with a function $time: O \to T$, where $time$ is injective. I'm assuming that $O$ is finite for this, and that $T$ does not have infinitely long chains. Find two elements $$t_{min} = \min(time(o)) \ o \in O$ \\ t_{max} = ...


2

You have $2$ variables $a_1$ and $v$ but only $1$ equation: $S_{25}=1275 \Rightarrow 25(a_1+12v)=1275$. Hence, infinite solutions are possible to your problem.


1

$$u_{11}=32$$ $$u_8 + u_{13} = 61$$ Therefore, $a + 10d = 32$ --- I and $(a + 7d) + (a + 12d) = 2a + 19d = 61$ --- II Now, 2 x eqn. I - eqn. II gives, $(2a + 20d) - (2a + 19d) = 32 * 2 - 61$ Therefore, $d=3$ Hence, using I, we get, a = 2. Thus, you now have the entire series there with you.


0

I get $6$ for the fifth term: $T(n)=\frac{378-272n+75n^2-7n^3}{3}$ gives the general term. $58=T(1)$ $26=T(2)$ $16=T(3)$ $14=T(4)$ $6=T(5)$


1

Use $a_n=a_1+d(n-1)$ $$\begin{cases} U_{11}=32 \\ U_8+U_{13}=61 \end{cases}$$ $$\begin{cases} U_1+10d=32 \\ U_1+7d+U_{1}+12d=61 \end{cases}$$


3

Hint: We know that $u_n = a + bn$ for some constants $a$ and $b$. So, from the information given, we know that $$ a + b(11) = 32\\ [a + b(8)] + [a + b(13)] = 61 $$ Using these equations, how can we find $a$ and $b$?


2

You get $2$ there, not $7/2$. $n^2 + 2n - 120 = 0$


0

I doubt your approximation. First I guess you missed a factor $2$ for $\ln 2$, the approximation should be $$f(x) =\ln(x+1)-\Psi(x+3/2)-2+\gamma+2\ln 2 \cdot$$ If you omit the $2$ you have $$f(0)= 0 -(2-\gamma-2\ln2) -2+\gamma+ \ln 2 = -4 + 2\gamma + 3\ln 2 \approx -0.766$$ while with the factor $2$ you have $$f(0)= 0 -(2-\gamma-2\ln2) -2+\gamma+ 2\ln 2 = ...


1

I'm not sure why there is a $(-2+\gamma+\ln 2)$, because $\left|\ln(x+1) - \Psi(x+\frac32)\right| < 0.0365$ itself is already satisfied. The approximation comes from the series expansion of $\exp\left(\Psi(x+\frac12)\right)$, which states, for $x > 1$, \begin{align} \Psi\left(x + \frac12\right) = \ln\left( x + \frac{1}{4!\cdot x} - \frac{37}{8\cdot ...


0

Let (1) $a_n = \frac{2^n}{3^{n+1}}$ Then it's easy to see that: (2) $a_{n+1} = \frac{2}{3} \cdot a_n$ Let's say we know that the sequence has a limit $L$. Then by letting $n$ go to infinity in (2) we get: (3) $L = \frac{2}{3} \cdot L$ So $L=0$. Now we just have to prove that the sequence has a limit. To do this it suffices to note that ...


4

$$\frac{2^n}{3^{n+1}}=\frac13\left(\frac23\right)^n$$ Now, what is $\;\lim\limits_{n\to\infty}x^n\;$ for $\;|x|<1;$ ?


2

Let $m> n$ be such that $l=m-n$ satisfies $$\left|\frac{\sum_{j=n}^{m-1} a_j^2}{l}-\rho\right|<\epsilon$$ for all $m>n$, where $\epsilon>0$ is chosen such that $\rho+\epsilon<1$. Then, an application of AM-GM inequality gives $$\prod_{j=n}^{m-1} a_j^2\le \left(\frac{\sum_{j=m}^n a_j^2}{l}\right)^l\le (\rho+\epsilon)^l$$ Since this is true for ...


1

[update] The corrected version in the question gives an easier result; the fixpoint to which the iteration proceeds is even expressible using the Lambert-W function. Let $a(0)=m-1$ , define always $b(k) = a(k)/m$, let $J=\exp(K)$ and iterate using the $b(k)$-version by $$ b(k+1) = J^{b(k)-1} $$ The fixpoint $t=\lim_{n \to \infty}b(n) $, if one exists, can ...


1

Remember that $a<b\implies a\le b$. This is because $a\le b$ means $a<b$ or $a=b$ as you already noted in the comment. If you are still confused, recall that "$p\implies q$" means $q$ is true whenever $p$ is true. And when $a<b$ is true, $a<b$ or $a=b$ is true. Hence, $a<b\implies a\le b$. For example, is "$0\le 1$" true?


0

Since nobody addressed the second part of the question $$ S(x) = \sum_{n=0}^{\infty} \frac{x^{3n+2}}{3n+2} \\ D_xS = \sum_{n=0}^{\infty} x^{3n+1} = x\sum_{n=0}^{\infty}\left ( x^{3} \right )^{n} = \frac{x}{1-x^3} \\ \text{Since} \ S(0) = 0, \ \text{we can find} \ S(x) \ \text{by solving} \\ S(x) = \int_{0}^{x}\frac{t}{1-t^3} \ \mathrm{d}t = ...


3

Limits don't necessarily preserve strict inequalities. For example, $1-1/n<1+1/n$, yet they have the same limit as $n$ goes to $\infty$.


0

Hint: If $\sum f_n$ converges uniformly on a set $E,$ then we must have $\sup_E|f_n| \to 0.$ Is that true here?


0

Hint 1: For $\lvert x\rvert<1$, use ratio test and see what you get. Hint 2: For $\lvert x\rvert>1$, use ratio test and see what you get.


2

Observe that $\;x^{2n}-1=(x^n-1)(x^n+1)\;$ , so in fact $$\frac{x^n}{x^{2n}-1}=\frac{x^n}{\left(x^n-1\right)\left(x^n+1\right)}=\frac12\left(\frac1{x^n-1}+\frac1{x^n+1}\right)$$ Try now to attack it from here.


6

If we set $f_n(x)=\frac{x^n}{x^{2n}-1}$ we have that $f_n(x)=- f_n\left(\frac{1}{x}\right)$, so it is enough to prove convergence over $x\in(-1,1)$ to have convergence over $\mathbb{R}\setminus\{-1,+1\}$. But if $|x|<1$ we have: $$ \frac{x^n}{1-x^{2n}} = x^{n}+x^{3n}+x^{5n}+\ldots $$ so: $$ \sum_{n\geq 1}\frac{x^n}{1-x^{2n}} = \sum_{n\geq 1} ...


1

We get an infinite preimage for any number other than 0,0.5,1. First note that if we have some "partial series" $\sum_{k=0} ^n (-1)^{k}A(k)^{-1}$ (with A a monotone function of k into the naturals), a necessary and sufficient condition for a continuation of the series to exist which converges to $x$ (that is, some $\tilde A\in R^{-1}(x)$ s.t. $\tilde A ...


1

As usual, take the $\;n\,-$ th roots test to the whole thing in absolute value: $$\sqrt[n]{\left|\frac{x^{3n+2}}{3n+2}\right|}=\frac{|x|^3\sqrt[n]{x^2}}{\sqrt[n]{3n+2}}\xrightarrow[n\to\infty]{}|x|^3$$ Thus, for $\;|x|^3<1\iff |x|<1\;$ we have convergence, and the interval of convergence is $\;[-1,1)\;$ . Another way: write ...


0

The series diverges for $x=1$, so its radius of convergence $R$ is smaller than or equal to $1$. If $x$ is such that $|x| < 1$, the series converges absolutely by comparison with the geometric series $\sum_{n\geq 0}{|x|^n}$. Then $R=1$. EDIT: I missed the sum part The series defines a smooth function $f:(-1,1) \to \mathbb{R}$ given by $f(x) = ...


0

As a generalization, if we subtract the first $m$ terms of the series for $e^x$ and divide by $x^m$, we get this: $f_m(x) =e^x-\sum_{k=0}^{m-1} \dfrac{x^k}{k!} = $ so $\begin{array}\\ \int_0^1 \dfrac{f_m(x)dx}{x^m} &=\int_0^1 \dfrac{\sum_{k=m}^{\infty} \dfrac{x^k}{k!}dx}{x^m}\\ &=\sum_{k=m}^{\infty} \dfrac1{k!}\int_0^1 \dfrac{x^kdx}{x^m}\\ ...


2

Note that ${2n!\over (n!)^2}={2n\choose n}$ is a binomial coefficient. You know that binomial coefficients, ${k\choose m}$ add up to $2^k$. In particular, you know ${2n\choose n}< 2^{2n}$. But then you see that it being right in the middle means it is larger than all the others, in particular it is larger than the average, which is ...


2

You are right, the series doesn't converge uniformly on $[0,+\infty)$, and you have correctly identified the reason for that. Now all that remains is to make a formal argument of that. We get that from the Lemma: Let $f_n \colon S \to \mathbb{R}$ be a sequence of functions such that the series $\sum\limits_{n = 0}^{\infty} f_n(x)$ is uniformly ...


0

Certainly $R^{-1}(1/3)$ is infinite, via the following sequence: $1/3$ $1/2 - 1/6$ $1/2 - (1/5 - 1/30)$ $1/2 - (1/5 - (1/29 - 1/870))$ Each expression is an alternating sum that adds to $1/3$, where the last term $N$ in the previous subset is replaced by $\{N-1, N(N-1)\}$. You can always repeat this unless your two largest elements of $A$ are ...


8

By Dirichlet's test, the series converges for $|a|=1$ unless $a=1$, since the partial sums of $\sum_na^n$ are bounded.


2

You proof looks basically correct to me. $\lim_{n\to\infty}\frac{e^x-1}{x}=1$ should be $\lim_{x\to 0+}\frac{e^x-1}{x}=1$. I would write (1) as $$(1)\int_0^1\frac{e^x-1}{x}dx=\int_0^1g(x)dx.$$


5

HINT: For (a) let $x$ and $y$ be distinct points in $[0,1]$, let $U$ be an open nbhd of $x$, and let $V$ be an open nbhd of $y$. What do you know about the sets $[0,1]\setminus U$ and $[0,1]\setminus V$? How big is the union of two countable sets? Is $\Bbb R$ countable? For (b), start by showing that a sequence $\langle x_n:n\in\Bbb N\rangle$ in ...


0

This is not an answer, just an illustration. Here is a graph of $\phi(x) = \sum_{n = -\infty}^\infty e^{-|x-n|}$ for $-5 \le x \le 5$.


4

Yesterday I said that this was Tauber's Theorem, the original tauberian theorem. It's not; Tauber's Theorem is the analogous result for Abel summability. This is Someone's Theorem. I'm going to give a proof of ST organized in what I feel is the "right" manner, deriving convergence from a sort of "maximal inequality"; then as a bonus we will see that one can ...


1

The theorem in question is that when $f'(x)$ is continuous and the integral converges, then $$\int_0^\infty \frac{f(ax) - f(bx)}x dx = \left(f(0) - f(\infty)\right)\log\left(\frac ba \right).$$ When $a = n^2 + 1, b = n^2$ and $f(x) = 2^{-x}$, the RHS reduces to $\log\left(1 - \frac 1{n^2+1}\right)$. Based on your later integrals, you wanted $a_n = n^2, b_n ...


0

The original series diverges, as you end-up subtracting then adding $1$. Anyway, by grouping the terms in pairs, you get a new series with the general term $$\left(\cos\left(\frac1{2n}\right)\right)^{2n}-\left(\cos\left(\frac1{2n+1}\right)\right)^{2n+1}\\ =\left(1-\frac1{2(2n)^2}+\cdots\right)^{2n}-\left(1-\frac1{2(2n+1)^2}+\cdots\right)^{2n+1}\\ ...


5

An infinite series $\sum\limits_{n} a_n$ will never converge if $a_n \not\to 0$. It doesn't matter if it is alternating or not.


8

The missing item is $42$ of course, and the sequence is: $$a_n=-\frac{41n^4}{12}+\frac{211n^3}{6}-\frac{1435n^2}{12}+\frac{1025n}{6}-83$$


0

General formula to find next term is~ $$x^3-1$$ $$1^3-1=0$$ $$2^3-1=7$$ $$3^4-1=26$$ $$4^3-1=63$$ $$5^3-1=124$$ Upcoming number in sequence is 215


0

It must be zero, and the sequence is, of course, $$5,10,0,50,122,0,5,10,0,50,122,0,5,10,0,\dots$$


0

The missing item is $42$ of course, and the sequence is: $$a_n=\frac{139n^4}{24}-\frac{797n^3}{12}+\frac{6413n^2}{24}-\frac{5023n}{12}+217$$


4

$2^2+1,3^2+1 , .., 7^2+1, 11^2+1$, seems like series of primes to me so i would guess 26


1

Additionnally, by applying Theorem $2$ (here), one obtains a closed form in terms of the poly-Stieltjes constants. Proposition. We have $$ \begin{align} \int_0^1\frac{x\,(\ln x-1)}{\ln(1-x)}\,dx= \ln 2+\gamma_1(2,0)-\gamma_1(1,0) \tag1 \\\\\sum_{n=1}^\infty\frac{\ln(n+2)}{n\,(n+1)}=\ln 2+\gamma_1(2,0)-\gamma_1(1,0) \tag2 \end{align}$$ where $$ ...


2

Probably the MVT helps: $$ f_n(x) = n \left( f'(\xi(x,n)) \frac{1}{n} \right) $$ for some $$ x < \xi(x,n) < x+ \frac{1}{n}. $$ Now use the fact that $\xi(x,n) \to x$ as $n \to +\infty$ and that $f'$ is uniformly continuous on $[a,b]$.


3

No this is not true, it would still be for allmost all $a$. As a counter example consider $a=1$ and $f$ given such that $f(n)=1$ but $\int_{n-1/2}^{n+1/2}f(x)dx = \frac{1}{n^2}$. If you need me to elaborate a bit more leave a comment.


2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} ...


2

HINT: Show that $$n^n e^{1 - n} < n!$$ for $n \in \mathbb{N}$ with $n \geq 2$. Does this tell you anything about the series?


5

By Frullani's theorem $$ \log(n+2)=\int_{0}^{+\infty}\frac{1-e^{-(n+1)x}}{xe^x}\,dx \tag{1}$$ hence by multiplying both sides by $\frac{1}{n(n+1)}$ and summing over $n\geq 1$ we get: $$ \sum_{n\geq 1}\frac{\log(n+2)}{n(n+1)}=\int_{0}^{+\infty}\frac{(1-e^{-x})\left(1-\log(1-e^{-x})\right)}{xe^x}\,dx\tag{2} $$ then, by substituting $x=-\log(u)$: $$ \sum_{n\geq ...



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