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