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A convergent example: $$\sum_{n=1}^{\infty} \left( \frac{1}{x+n} + \frac{1}{x-n} \right) = \pi\cot{\pi x}-\frac{1}{x},$$ and the convergence is uniform on closed intervals not containing integer points. Clearly every function in the sum tends to $0$ as $x \to \infty$, but the entire sum does not. Edit: An even better example, perhaps: set $$s_n(x) = ... 4 This is not true. Consider s_n(x) = 1/x. We have \lim_{x \to \infty} s_n(x) = 0 = t_n. Clearly, \sum_{n=1}^{\infty} t_n =0 < \infty. However,$$S_n(x) = \sum_{k=1}^n s_n(x) = \dfrac{n}x$$and \lim_{n \to \infty} S_n(x) doesn't exist. 4 As others have mentioned, this is not true in general. But there is a very important case where it is true: power series. Abel's theorem says that if S(x) = \sum_{n=0}^\infty a_n x^n and S(1) converges, then \lim_{x \to 1-} S(x) = S(1). Since you're interested in x \to \infty, this will require a transformation to apply to your case. Thus take ... 4 The error is in the third line. The partial sum of a geometric series is valid only if the common ratio is not 1. 3 You can also find examples in which all series are convergent, yet you cannot pass to the limit. Try this:$$ s_n(x)= \begin{cases} 1, & n=[x] \\ 0, & \text{otherwise} \end{cases} $$Here [x] is the floor or ceiling function (choose the one you like most, it does not matter). To guarantee that you can pass to the limit under the sum you need ... 2 No. Let s_n(x)=\frac 1x, s_n=0. Then S=\sum_{n=1}^\infty s_n trivially converges. However, S(x)=\sum_{n=1}^\infty s_n(x) is clearly divergent, since the summand is a constant (with respect to n) greater than 0. 2 Here you have a sufficient condition (from the discussion in the comments) for your sequence to be convergent: Suppose that |a_n-a_{n+1}|\leq c_n, where \{c_n\} is such that$$ \sum_{n=1}^{\infty}c_n<\infty. $$We define the sequence of positive numbers C_n=\sum_{k=1}^nc_n, which is increasing. Since it converges, it is also Cauchy. Now we prove ... 2 You are just lacking a way to express the number of elements in the sequence, it seems. You can use \#A or |A| or state that the sequence has length N. Then append the suffix ^{th}. The Python programming language uses the following indexing convention: elements are numbered 0 to N-1, but negative indexes can be used. The element A_{-1} is the ... 2 It converges conditionally and not absolutely because \displaystyle \sum_{n=2}^\infty \left|(-1)^n\cdot \dfrac{1}{\ln n}\right| = \displaystyle \sum_{n=2}^\infty \dfrac{1}{\ln n} \geq \displaystyle \sum_{n=2}^\infty \dfrac{1}{n} = \infty, and you've done the first part that series is convergent by the alternating series test. 2 I would separate the constant and k terms like this: \begin{array}\\ A(n) &=\sum_{k=0}^{n-1}\binom{n}{k}\frac{(-1)^{n-k}[2(n-k)-1]-1}{2}A(k)\\ &=\sum_{k=0}^{n-1}\binom{n}{k}(-1)^{n-k}(n-k)A(k)-\sum_{k=0}^{n-1}\binom{n}{k}\frac{(-1)^{n-k}+1}{2}A(k)\\ ... 2 By ratio test$$\lim_{n\rightarrow\infty}\left|\frac{x^{n+1}n!\left(n+1\right)^{3}}{\left(n+1\right)!x^{n}n^{3}}\right|=\lim_{n\rightarrow\infty}\left|\frac{x}{n+1}\right|=0 $$so the radius of convergence is \infty and the interval of convergence is \left(-\infty,\infty\right). 1 Suppose that a_n=a_0+a'n and b_n=b_0+b'n are the arithmetic progressions. If there is at most one common term, then obviously it is not an arithmetic progression, and if a' and b' have different signs (one is increasing and one is decreasing), then there are only finitely many common terms so it is not an arithmetic progression. Otherwise, suppose ... 1 Recall the species of set partitions with the constituents marked which is$$\mathfrak{P}(\mathcal{U}\mathfrak{P}_{\ge 1}(\mathcal{Z}))$$which gives the generating function$$G(z, u) = \exp(u(\exp(z)-1)).$$It follows that the exponential generating function of Bell numbers is given by$$G(z) = \exp(\exp(z)-1).$$Suppose we are trying to compute ... 1 You are correct that the alternating series gives you the answer you want about convergence (although not about absolute convergence). In general, these tests are phrased as: "If such-and-such a condition is true, then the series converges." If you like, think of this in terms of predicates: you have a statement of the form P \implies Q. However, given ... 1 We have \sqrt[3]{n^3+an}=n\sqrt[3]{1+\frac{a}{n^2}}. We expand the latter using Newton's Binomial Theorem to get$$\sqrt[3]{n^3+an}=n\sum_{k\ge 0} {1/3 \choose k} \left(\frac{a}{n^2}\right)^k=n(1+\frac{a}{3n^2}-\frac{a^2}{9n^4}+O(n^{-6}))$$Repeating, we have \sqrt{n^2+3}=n\sqrt{1+\frac{3}{n^2}}, so$$\sqrt{n^2+3}=n\sum_{k\ge 0}{1/2 \choose ...

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How about $a_n = \frac{1}{n}$ and $x_n = n$? Then $\sup a_nx_n = 1$ do we have $0 \times +\infty =1$, but then, what about $b_n = \frac{2}{n}$

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First we can observe that $n^{1/n}\to1$. In fact $n^{1/n}=e^{\frac{\ln(n)}{n}}$ and $\frac{\ln(n)}{n}\to0$. Therefore the denominator tends to $1$ while the numerator tends to $+\infty$.

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Taylor series don't always converge. The Taylor series method can be very good if the interval of integration is inside the radius of convergence of the series, otherwise it may not converge at all. But as Yves Daoust mentioned, you should be comparing this to more sophisticated numerical methods, not to Riemann sums. In practice, Riemann sums are almost ...

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Here is another condition which is sufficient to allow you to interchange the limit and the sum. Let $$S(x) = \sum_{n=1}^\infty s_n(x), \qquad S_m = \sum_{n=1}^m \lim_{x \rightarrow \infty} s_n(x) = \sum_{n=1}^m s_n, \qquad S_m(x) = \sum_{n=1}^m s_n(x).$$ We know that $S_m \rightarrow S$ and want to know if also $S(x) \rightarrow S$. If each of ...

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$$\frac{a_{n+1}}{a_n}=\frac{8(n+1)!/(n+1)^{n+1}}{8n!/n^n}=\frac{n^n}{(n+1)^n}\to \frac{1}{e}$$ as $n \to \infty$. Since$1/e<1$, then the series converges!

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Perhaps the sequence $a_1,a_3,a_5,...$ is monotone, and also $a_2,a_4,a_6,...$

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$\textbf{HINT:}$ Try checking the secuence $\{a_{2k}\}_{k\in\mathbb{N}}$ and $\{a_{2k-1}\}_{k\in\mathbb{N}}$. One of them is crecent and the other decrecent. But it converges iff $\limsup=\liminf$

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You can show that $\arctan'(x) =\frac1{1+x^2}$ from the functional equation $\arctan(x)-\arctan(y) =\arctan(\frac{x-y}{1+xy})$ (gotten from $\tan(x+y) =\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}$). $\begin{array}\\ \arctan(x+h)-\arctan(x) &=\arctan(\frac{(x+h)-x}{1+(x+h)x})\\ &=\arctan(\frac{h}{1+(x+h)x})\\ \end{array}$ From $\sin(x) \approx x$ ...

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[Comment becomes answer] Let $a_n=n!+2$ for $n=1,2,\dots$. Then for any $k\ge0$, and any $n\ge k+2$, we see that $k+2$ is a proper divisor of $a_n+k=n!+k+2$, since $k+2$ divides $n!$. Thus, the sequence $a_n+k$, $n=1,2,\dots$, contains at most $k+1$ primes.

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