# Tag Info

2

This is true because if $(a_{2n})$ and $(a_{3n})$ are convergent then all their subsequences converge to the same limit. In particular consider the sequence $(a_{6n})$. Then $\lim_\limits{n\to \infty}a_{2n}=\lim_\limits{n\to \infty}a_{6n} =\lim_\limits{n\to \infty}a_{3n}$.

6

It's true, for $\{a_{6n}\}$ is a subseqeunce of $\{a_{2n}\}$ ($a_{6n} = a_{2(3n)}$) and a subseqeunce of $\{a_{3n}\}$ ($a_{6n} = a_{3(2n)}$). Thus $$\lim_{n\to \infty} a_{2n} = \lim_{n\to \infty} a_{6n} = \lim_{n\to \infty} a_{3n}.$$

2

This is false. Consider the sequence $a_{n}=\frac{1}{i^2}$ if $2^{i}=n$ and $a_{n}=0$ otherwise. $a_{n}\in \ell^{1}$ because summing it is (padded with zeros) the same as $\sum \frac{1}{n^2}$ which of course converges. But $a_{n}n=\frac{2^{i}}{i^2}\to \infty$ whenever $n$ is a power of $2$. So $\limsup na_{n}=\infty$.

0

This is problem number 17 from page 35, Examples IV. b. The original solution of this problem from page 10 of Solutions of the Examples in Higher Algebra (LaTeX Edition) by Hall and Knight is

0

For $a > 1$: By the limit definition of $\lim_{n\to\infty}a_n=a$, there is an $N \in \mathbb N$ with $$n > N \Rightarrow |a_n-a| < \epsilon.$$ for all $\epsilon > 0$. Let $\epsilon = \frac{a-1}{2}$, then $1 < a-\epsilon$, and therefore, for all $n > N: a_n > 1$. From this we know, again for all $n > N$, that $a_n^2 > a_n$. Thus, ...

0

Suppose $(a_n)$ is a sequence and $\lim_{n\to\infty} a_n = a$. Then you know by algebra of limits that $\lim_{n\to\infty} a_n^2 = a^2$. - Fact 1. This is a basic fact, so you can look it up on the internet, but I think you should have been taught that. Now suppose $a>1$. So $a-1 = \varepsilon, \exists \varepsilon > 0$. Then there exists an $M_1$ such ...

0

This is problem number 7 from page 34, Examples IV. b. The complete solution of this problem from page 9 of http://www.amazon.com/Solutions-Examples-Higher-Algebra-LaTeX/dp/150876350X/ is

1

$$\lim_nb_n=\max\{a,a^2\}=\begin{cases}a^2 & \mbox{ if } a>1\\ 1 &\mbox{ if } a=1\end{cases}.$$ Added to the proof Since $$\max\{x,y\}=\frac{x+y+|x-y|}{2} \quad \forall x,y\in \mathbb{R}$$ we have \begin{eqnarray} |\max\{x_1,y_1\}-\max\{x_2,y_2\}|&=&\frac12\left|x_1+y_1+|x_1-y_1|-x_2-y_2-|x_2-y_2|\right|\\ ...

0

Your argument uses the axiom of countable choice, which says that every countable family of non-empty sets has a choice function. This question (one of those to which you linked) shows that some amount of choice is needed. The axiom of choice is usually not mentioned in such arguments simply because it’s generally taken for granted in doing topology.

2

Let $\omega=\exp\frac{2\pi i}{3}$. Then: $$\sum_{n\geq 1}\frac{1}{n^4+n^2+1}=\frac{1}{i\sqrt{3}}\left(\sum_{n\geq 1}\frac{1}{n^2-\omega}-\sum_{n\geq 1}\frac{1}{n^2-\omega^2}\right)=\color{red}{-\frac{1}{2}+\frac{\pi\sqrt{3}}{6}\tanh\frac{\pi\sqrt{3}}{2}}$$ since we can deal with series like the ones appearing in the middle term through the logarithmic ...

0

Are you sure that the denominator is $f''(b)$? Because I see almost the same formula in the proof of quadratic convergence for the Newton's method, see the expression after the phrase "one finds that". Although, there is also this square...

0

You could use the Calabrese criteria (see the references) to approximate the sum of alternating series: References: http://ecademy.agnesscott.edu/~lriddle/apcalculus/approxSeries.pdf

3

Use the Borel-Cantelli lemma. Essentially, for the series to converge almost surely, the probability X is bigger than 0 necessarily has to converge to zero, which means that P({X=0 i.o.}) = 1. Clearly that's not the case since P({X≠0}) > 0 and constant for all X. Therefore, the sum of the probabilities diverges. And, because the sum of the probabilities ...

1

$$\frac 1 9 + \frac 1 {10} + \frac 1 {11} + \cdots + \frac 1 {16}\ \ge\ \frac 1 {16} + \frac 1 {16} + \frac 1 {16} + \cdots + \frac 1 {16} = \frac 1 2$$ $\ldots$ and similarly for the such sums.

0

How many terms in each expression? Replace each one with the minimum term in the expression and ....

1

Although this may not be needed as of now; but I thought about sums of quadratic sequences myself and I managed to derive a general formula for it, so I might as well post it here: Where is the number of terms to compute, is the starting term, is the first difference and is the constant difference between the differences. I use the subscript to denote ...

1

Yes, it has a minimum number, if $a_n \rightarrow \infty$ then as you said for any $K > 0$ there is an $N$ such that for all $n \geq N$ implies $a_n > K$. Then the minimum is $\min \{a_1, \dots, a_N\}$

4

If $a_n \to a >0$, then $\log a_n \to \log a$ by continuity of log (and $\log a$ is finite). Also, $\frac{1}{n} \to 0$, so $\frac{1}{n} \log a_n \to (0) \log a = 0$. By continuity of exponentiation, we have $(a_n)^{1/n} = e^{\frac{1}{n} \log a_n } \to e^0 = 1$, as desired.

4

\begin{align}\sum\limits_{i=1}^n (7+6i)^2 &= \sum\limits_{i=1}^n (49 + 84i + 36i^2)\\& = \sum\limits_{i=1}^n 49 + \sum\limits_{i=1}^n 84i + \sum\limits_{i=1}^n 36i^2\\ &=49\sum\limits_{i=1}^n 1 + 84\sum\limits_{i=1}^n i + 36\sum\limits_{i=1}^n i^2\end{align} What do you know about $\sum\limits_{i=1}^n 1$? What about $\sum\limits_{i=1}^n i$? ...

1

You don't even need calculus to prove that $u_n = \frac{4}{(\ln(n))^2}$ is monotonically decreasing, for large enough $n$. The argument rests on three facts: that $\ln(n)$ is increasing; that is, $\ln(n) < \ln(n + 1)$, that for positive $a, b$ both greater than $1$, if $a > b$ then $a^2 > b^2$ that for positive $a, b, c$, the inequality ...

0

You have to prove that: $$\frac{1}{(\ln n)^2}\gt\frac{1}{(\ln (n+1))^2}$$ The inequality is very simple.

0

The Alternating Series Test basically says if a series is monotonically decreasing (absolute value of terms is decreasing) and $\lim_{n\to \infty}$ of the series is equal to $0$, then the series converges. You are going in the right direction. $$\frac{4}{(\ln n)^{2}}$$ decreases monotonically and, as you noted, $\lim_{n\to \infty}$ of this series equals ...

2

You have $$e^{-y(s)}y'(s)=1,$$ hence $$\left(e^{ -y(s )}\right)'=-1.$$ Integrate with respect to $s$ from $0$ to $x$: $$e^{-y(x)}-e^{-y(0)}=0-x,$$so $$y(x) =-\ln(1-x).$$ This should be sufficient to find the parameter $a$.

0

For simplicity of derivation, let $c = \cosh 1$ and $s = \sinh 1$, we have \begin{align} \sinh(x+1) - \sinh x &= (c - 1) \sinh x + s \cosh x \tag{*1a}\\ \cosh(x+1) - \cosh x &= (c - 1) \cosh x + s \sinh x \tag{*1b} \end{align} $s \times (*1a) - (c-1) \times (*1b) \implies$ \begin{align} & s \text{LHS}(*1a) - (c-1)\text{LHS}(*1b)\\ =\; ... 0 Assuming x,m\in\mathbb{N}\setminus\{0\}, let we replace x-1 with n and m-1 with s. Then:\sum_{i=1}^{n}\left(1-\frac{1}{2^i}\right)^s = \sum_{i=1}^{n}\sum_{k=0}^{s}\binom{s}{k}\frac{(-1)^k}{2^{ik}}=\sum_{k=0}^{s}\binom{s}{k}(-1)^k\frac{1-2^{-kn}}{2^k-1}$$but I do not see any particular reason for considering the last sum "prettier" than the ... 5 First, let us recall following integral representation:$$\frac{1}{k(k+1)} = \int_0^1 \int_0^y x^{k-1} dx dy$$We have$$ \sum_{k=1}^n \frac{1}{k(k+1)} = \int_0^1 \int_0^y \left( \sum_{k=1}^n x^{k-1} \right) dx dy = \int_0^1 \int_0^y \left( \frac{1-x^n}{1-x} \right) dx dy\\ = \int_0^1 \int_x^1 \left( \frac{1-x^n}{1-x} \right) dy dx = \int_0^1 (1-x^n) dx ...

0

Break the argument: $$\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$$ Then: $$\sum_{k=1}^n\frac{1}{k(k+1)} = \sum_{k=1}^n\frac{1}{k}-\sum_{k=1}^n\frac{1}{k+1}$$ Let's pose $$H_n = \sum_{k=1}^n\frac{1}{k}$$ Then you have: $$\sum_{k=1}^n\frac{1}{k(k+1)} = H_n - \sum_{k=2}^{n+1}\frac{1}{k}=\\=H_n-\left(\sum_{k=1}^{n+1}\frac{1}{k} - \frac{1}{1}\right) = ... 0 By partial summation we have$$\sum_{k\leq n}\frac{1}{k\left(k+1\right)}=\frac{H_{n}}{N+1}-\sum_{k\leq n-1}H_{k}\left(\frac{1}{k+2}-\frac{1}{k+1}\right) $$where H_{k}  is the k-th  armonic number. This sums have a closed form$$\sum_{k\leq n-1}\frac{H_{k}}{k+1}=\frac{1}{2}\left(H_{n}^{2}-H_{n}^{(2)}\right) \sum_{k\leq ...

2

It is the logarithm of the Wallis product, hence such a series converges to $\color{red}{\log\frac{\pi}{2}}$.

1

One neat way you can do this is via generating functions but I think all it really does is obfuscate the fact that there is a telescoping happening. Nevertheless here's how it works. Let $s_n$ be the sum in question. Let $f(x)$ be the generating function of the the sequence in the sum, that is $$f(x) = \sum_{k=1}^\infty \frac{x^k}{k(k+1)}$$ To get at $s_n$ ...

0

Using term-wise derivation, we define the auxiliary sum $$e^{ix}=\sin'(x)+i\sin(x)=\sum_{k=0}^\infty\frac{(ix)^k}{k!}.$$ By a straightforward manipulation of this sum using the Binomial theorem, ...

1

For the last series, since the second is convergent it suffices to check that the series $\sum_{n\geqslant 1}b_n^2\mathbb E\left[X_1^2\mathbf 1\{|b_nX_1|\leqslant 1 \}\right]$ is convergent. To this aim, we start writing that $$b_n^2\mathbb E\left[X_1^2\mathbf 1\{|b_nX_1|\leqslant 1 \}\right]\leqslant \int_0^1 t\mathbb P\{|b_nX_1|\geqslant t\}\mathrm dt.$$ ...

0

There is one more great mistake: The limit in the root test does not equal the sum of the series!!

1

$$\lim_{n \to \infty}\left(1+\frac{3}{n}\right)^n=e^3$$ Moreover the root test is applied on the $n$th term. You don't find the sum of the series like that.The test is only for convergence or divergence.

1

The limit is of the form: $$\lim_{n\to\infty}\left(1+\frac{m}{x}\right)^{nx}~\textrm{with }m=3,n=1$$ We know that the limits in these form have the solution $e^{mn}$. Refer to this link for all the limit-related identities. For the problem here, the limit is $e^3$.

1

$$L=\lim_{x\rightarrow \infty }(1+\frac{3}{n})^{n^2/n}=\lim_{x\rightarrow \infty }(1+\frac{3}{n})^n$$ after that take the log for both sides $$\log(L)=n\log(\lim_{x\rightarrow \infty }(1+\frac{3}{n}))$$ by using the Lopital Rule the limit is $e^3>1$ so the series is diverge

2

By partial summation we have $$\sum_{k=2}^{n}\log\left(2-e^{1/k}\right)=n\log\left(2-e^{1/n}\right)-2\log\left(2-e^{1/2}\right)-\int_{2}^{n}\left\lfloor t\right\rfloor \frac{e^{1/t}}{t^{2}\left(2-e^{1/t}\right)}dt$$ where $\left\lfloor t\right\rfloor =t-\left\{ t\right\}$ is the floor function and $\left\{ t\right\}$ is the sawthoot function. Then ...

1

Let $\mathbb{R}^{*} = \mathbb{R} \cup \{ \infty \}$ and $f : \mathbb{R}^{*} \to \mathbb{R}^{*}$ be the function $f(u) = \frac{2}{2-u}$, one can check that $$f^{\circ 4}(u) = f(f(f(f(u)))) = u$$ From this, we see unless $f(u) = u$, the sequence $( u_n )$ defined by $$u_n = \begin{cases} u, & n = 0\\ f(u_{n-1}), & n > 0\end{cases} \quad\iff\quad ... 0 Let f(x) = \frac{2}{2-x} and consider differencies |x - f(x)|. If u_n converges, it should be a Cauchy sequence. If we manage to prove that |x - f(x)| cannot be as close to zero as we want, then the series cannot converge. Since we are only interested in how close it can be to zero, sign does not matter and we can consider g(x) = x - f(x) = x - ... 0 The equality is due to the fact that:$$\sum_{n=1}^\infty\frac{1}{(n+2)^6} = \frac{1}{3^6} + \frac{1}{4^6} + \frac1{5^6}+\dots$$and$$\sum_{n=3}^\infty\frac{1}{n^6} = \frac{1}{3^6} + \frac{1}{4^6} + \frac1{5^6}+\dots$$The equations have the same right-hand side, so their left-hand sides are equal as well. 0$$\sum_{n=1}^\infty\frac{1}{n^6}=\frac{1}{1^6}+\frac{1}{2^6}+\frac{1}{3^6}+\frac{1}{4^6}+\dots\sum_{n=1}^\infty\frac{1}{(n+2)^6}=\frac{1}{3^6}+\frac{1}{4^6}+\dots=\left(\sum_{n=1}^\infty\frac{1}{n^6}\right)-\frac{1}{1^6}-\frac{1}{2^6}$$Do you understand it now? 4 First you should show that u_n is well defined for all n if u_0 \notin \{0,1,2\}. Hint: induction. If the sequence converged to u\neq 2 then you could take the limit of both sides to get u = \frac{2}{2-u} which has solutions u=1\pm i. Since sequences of real numbers can't converge to 1\pm i, it couldn't possibly be that u_n converges unless ... 5 if not, let u_n\to u as n\to \infty, then you will have$$u=\frac{2}{2-u}$$Does it have a solution? 1 For n\geq 1, let P(n) denote the following statement:$$ P(n) : \sum_{k=1}^n k^3=\left(\frac{n(n+1)}{2}\right)^2. $$(a): P(1) is the statement that \sum_{k=1}^1 k^3 = \left(\frac{1(1+1)}{2}\right)^2. (b): P(1) is true because \sum_{k=1}^1 k^3 = 1 = \left(\frac{1(1+1)}{2}\right)^2. (c): Fix some \ell\geq 1 and assume that P(\ell) is true ... 4 A great starting point on these questions is to "ignore all the small pieces" and see what happens. For instance, for large n, the +1 in the denominators won't really matter. So the first really looks like$$ \sum_{n \geq 1} \frac{1}{n^2}$$and the second looks like$$ \sum_{n \geq 1} \frac{1}{n}.$$Some calculus books have a so-called "limit comparison" ... 1 No. The inductive hypothesis is \sum\limits_{k=1}^n k^3 = {({{\frac{{n(n+1)}}{2}}})}^2 , which we assume is true. From there:$$ \sum\limits_{k=1}^n k^3 +(n+1) = {({\frac{n(n+1)}{2}})}^2+(n+1) $$Edit - the above has a typo, it should be:$$ \sum\limits_{k=1}^n k^3 +(n+1)^3 = \left ({\frac{n(n+1)}{2}}\right )^2+(n+1)^3 $$Try to see if you can go on from ... 2 You would use the first one. The idea is to suppose you have an operator r that takes each element of the sequence to the next one--that is, rs_n=s_{n+1} for all n. Then$$r^2s_n=2rs_n-s_n$$for all n by the recurrence, and so$$(r^2-2r+1)s_n=0$$for all n. Since s_0=1, then it follows that$$r^2-2r+1 is the zero operator (sends everything to ...

1

Also $\frac{n\cdot(n-1)\cdots 2\cdot 1}{n\cdot n \cdots n} \le \frac{2\cdot 1}{n\cdot n} = \frac{2}{n^2}$ for large $n,$ if that helps

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