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8

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


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} ...


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 ...


4

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


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 $\infty$.


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 ...


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 $$(1)\int_0^1\frac{e^x-1}{x}dx=\int_0^1g(x)dx.$$


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.


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 ...


2

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 $7/2$. $n^2 + 2n - 120 = 0$


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 ...


1

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

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 ...


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?


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 ...


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

[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 ...



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