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4

This means that $g(n) = (f(n) + h(n))^n$ for some function $h$ such that $h(n) = o(1/n)$.

3

This expression means that $g(n)=\left(f(n)+h(n)\right)^n$ where $h(n)$ is some (unspecified) function that is $o(1/n)$.

2

The notation $f(t) \sim g(t)$ (which means that $\lim_{t \to 0^+} f(t)/g(t) = 1$) would only tell you that the first nonzero terms in the asymptotic expansions for $f(t)$ and $g(t)$ are equal (presuming you already know that the expansions are of the same form). As far as I know there is not a common notation which means that two functions share the same ...

2

You are interested in the integral $$f(x) = \int_1^x y\sin\!\left(\frac{2\pi (y-1) x}{y}\right) dy,$$ and this is very close to the integral $$g(x) = \int_0^x y\sin\!\left(\frac{2\pi (y-1) x}{y}\right) dy,$$ which is (basically) expressible in closed form. To see this, first make the substitution $y = ux$ to get $$g(x) = x^2 \int_0^1 u ... 1 Let U(k)=\frac{T(k)}{k!}. Then$$U(n)=U(n-1)+\frac{n}{(n-1)!}=U(n-1)+\frac1{(n-1)!}+\frac1{(n-2)!}\\ <U(1)+2e-1\\ T(n)<2e(n!)$$1 You know that$$\lim_{n\to\infty}n^2-n=\infty$$By definition this means for all N>0 there exists M(N)>0 such that n>M implies n^2-n>N. But then let N>\log_2(c). Then$$2^{n^2-n}>2^N>2^{\log_2(c)}=c.$$1 This isn't true. For instance, suppose your original sequence is in increasing order. Then k_\max=0, but k_{\max \text{sorted}}>0 unless all the terms are equal. However, it is true (with a \leq instead of <) if you assume either the first or the last number is already sorted, i.e. either s_1=\tilde{s}_1 or s_k=\tilde{s}_k. Let us ... 1 This is a very basic question that follows trivially via summation by parts. We have that $\sum_{n \leq x} \frac{d_{sf}(n)}{n} = \frac{1}{x} \sum_{n \leq x} d_{sf}(n) + \int_{1}^{x} \frac{1}{t^2} \sum_{n \leq t} d_{sf}(n) \, dt.$ Using the expression $\sum_{n \leq x} d_{sf}(n) = A x \log x + B x + E(x),$ where the error term E(x) satisfies E(x) = ... 1 The error comes from replacing \binom{k}{2} with k^2/2. Leave it as k(k-1)/2. Then the exponent on the 2, after dividing by k-2 becomes$${\frac{k(k-1)}{2}\cdot\frac{1}{k-2}}= (\frac{k}{2}-\frac{1}{2})\cdot(1 +\frac{2}{k} + O(1/k^2)) = \frac{k}{2} + \frac 12 + o(1).$$If you do Erd\H{o}s's original argument and leave \binom{k}{2} as k(k-1)/2 you ... 1 You need to show that there is some M>0 and some N>0 such that \forall n\geq N, n^d\leq M(1+\epsilon)^n. Pick M=1 and let's find N:$$ n^d\leq(1+\epsilon)^n\iff d\log n\leq n\log(1+\epsilon)\iff\frac{\log n}{n}\leq\frac{\log(1+\epsilon)}{d}. $$Consider the last inequality. At n=e, the LHS evaluates to \frac{1}{e} and for n\geq e, the ... 1 n is a real if we suppose l_n= \sum_{0\le k \le n} \frac1{k+1} So l_n = 0 for n <0, l_n = 1 for n \in [0,1[, ... Remarks: 1/ If C\ge 0, -C \le 0 2/  n \ln(n) >0 if n > 1. proof: ==> If n> -1, as \ell_n \ge 0 for all n, then (n+1)\ell_{n \pm 1} \ge 0 \ge -C. <== if (n+1)l_{n\pm 1} \ge 0 , as \ell_n \ge 0 ... 1 The upper O-bound is trivial. To establish the lower bound, consider the following limit and apply L Hospital rule repeatedly$$\lim_{x\to\infty}\frac{\lg^{k}2x}{\lg^{k}x}$$Note that$$\dfrac{d}{dx}\left(\lg2x\right) =\dfrac{d}{dx}\left(\lg x\right)$$1 You don't need to calculate anything. Just note: In a polynomial the fastest growing (decreasing) term will always be the monomial of highest order. For a general polynomial a_0+a_1x+...+a_nx^n this is the a_nx^n. Meaning: For large values of n this term will dominate all the others so you could also just have a look at the highest order term in ... 1 Hint. You may compare your sums with related integrals.$$ \int_1^n (\ln x)^sdx-(\ln (n+1))^s\leq\sum_{k=1}^n (\ln k)^s \leq\int_1^n (\ln x)^sdx,\quad n\geq1. $$Then observe that, as n \to \infty,$$ \begin{align} \int_1^n (\ln x)^s\:dx&=n\:(\ln n)^s-s\int_1^n (\ln x)^{s-1}\:dx\sim n\:(\ln n)^s. \end{align}  Similarly, as $n \to \infty$, ...

1

Using integrals which needed to be guessed looked overly complicated for this problem, i asked a friend at work and we believe that we came up with an answer that does not involve integrals: $\sum_{k=1}^nlg^sk = \sum_{k=1}^{\lfloor n/2\rfloor}lg^sk+\sum_{k=\lfloor n/2\rfloor}^nlg^sk\ge$ $\sum_{k=1}^{\lfloor n/2\rfloor}0+\sum_{k=\lfloor ... 1 For the sequence$b_n$constructed in the comments, it is not difficult to see that$0\le b_n\le 1/2$(We can use induction to prove the second inequality and the first one is obvious). Then, by Bolzano-Weierstrass theorem there is a subsequence which converges. Let$\{b_{n_k}\}$be such a subsequence. Let the limit be denoted by$b\$. Then the recursion ...

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