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6

Formally speaking, neither is correct. Sequences are functions, and the elements of $M$ are not [usually] functions from $\Bbb N$ to $M$ itself (although that is known to be possible). What is correct is that "$(x_n)$ is a sequence such that $x_n\in M$ [for all $n$]". To me, then, both notations abuse about the same the correct way of phrasing, so both are ...


6

Neither $(x_n)\subset M$ nor $(x_n)\in M$ should be used, since neither is correct. The fact that we say "sequence in $M$" does not mean we should use the symbol for element in a set. There are clear and precise definition for the meaning of $\in $ and $\subset$ and they dictate what the correct usage is. Generally speaking, a sequence $(x_n)$ of elements in ...


4

The hypothesis is wrong. Let $f$ be the following sequence: $1,2 ; 3,5,4,6; 7,9,11,8,10,12; \cdots$ where each block has two more numbers than the previous block and goes through the odd numbers in order before the even ones. Clearly $f(n) \in n + O(\sqrt{n})$ as $n \to \infty$. Now consider the following series: $( \frac{1}{1} - \frac{1}{1} ) + ...


3

The binomial theorem states that: $$(x+y)^n=\sum_{k=0}^{n}{{n}\choose{k}}x^ky^{n-k}$$ Using this, with $y=1$, we see that the RHS is equal to: $$\left(1+\frac{g}{100}\right)^t$$ This is perfectly logical. Each year the population is multiplied by $1+g\%$. So after $t$ years it's multiplied by $(1+g\%)^t$. Now to find $t$ you just solve $2=(1+g/100)^t$, ...


3

Using $$x-1 \le [x] \le x,$$ you can safely remove the floor function signs without changing the limit. From there you should be good.


3

$$a=\dfrac1{x-3y},b=\dfrac1{x-y},c=\dfrac1{x+y},d=\dfrac1{x+3y}$$ $\implies a+d=\dfrac{2x}{x^2-9y^2}$ and $b+c=\dfrac{2x}{x^2-y^2}$ $\implies a+d>b+c$ as $9y^2>y^2$


2

The correct option is $(C)$. We have $$\frac 1a+\frac 1c=\frac 2b\quad\text{and}\quad \frac 1b+\frac 1d=\frac 2c$$$$\Rightarrow c=\frac{ab}{2a-b}\quad\text{and}\quad d=\frac{ab}{3a-2b}$$ Here, since $c,d$ are positive, we have $2a-b\gt 0,3a-2b\gt 0$. By the way, we have ...


2

Recall the difference of two cubes factorisation: $$a^3 - b^3 = (a - b)(a^2 + ab + b^2).$$ Now consider the above with $a = (n^3 + 1)^{1/3}$ and $b = n$. Then, $$((n^3 + 1)^{1/3} - n)((n^3 + 1)^{2/3} + n(n^3 + 1)^{1/3} + n^2) = n^3 + 1 - n^3 = 1.$$ Rearranging, $$(n^3 + 1)^{1/3} - n = \frac{1}{(n^3 + 1)^{2/3} + n(n^3 + 1)^{1/3} + n^2}.$$ Try applying the ...


2

$$x-1\leqslant \left \lfloor x \right \rfloor< x\\2x-1\leqslant \left \lfloor 2x \right \rfloor< 2x\\3x-1\leqslant \left \lfloor 3x \right \rfloor< 3x\\...\\\\nx-1\leqslant \left \lfloor nx \right \rfloor< nx$$ now sum of them $$(x+2x+3x+..+nx)-(1+1+1..1)\leq \left \lfloor x \right \rfloor+\left \lfloor 2x \right \rfloor+...+\left \lfloor nx ...


2

Hint: Evaluate $$\frac1{a+n}+\frac1{a-n}\,.$$


1

HINT: Note that for the substitution $n \to -n$ the series is unchanged. Thus, the sum over non-zero integers is twice the sum over the positive integers only. Can you finish now?


1

Hint: The $n^{\text{th}}$ term of the arithmetic series is given by $t_n=t_1+(n-1)d$, where $t_1$ is the first term of the series and $d$ is the difference between successive terms. The sum of the first $n$ terms of an arithmetic series is then given by $S_n = \frac{n(t_1 + t_n)}{2}$. Finally, the sum of the first six terms minus the sixth term is of ...


1

Note that $$\sum_{k=0}^n a_k=\sum_{k=0}^{j-1} a_k+\sum_{k=j}^n a_k$$ That first term on the right hand side does not depend on $n$ and is just a constant. Therefore $$\begin{align} \sum_{k=0}^\infty a_k&=\lim_{n\to\infty}\sum_{k=0}^n a_k \\ &=\lim_{n\to\infty}\left(\sum_{k=0}^{j-1} a_k+\sum_{k=j}^n a_k\right) \\ &=\sum_{k=0}^{j-1} ...


1

Hint: We know power series are absolutely convergent in their radius of convergence, hence $$\left|\sum_{n=1}^\infty\left[(n^3+1)^{1/3}-n\right]x^n\right|\leq\sum_{n=1}^\infty 3nx^n,$$the latter having a radius of convergence $1$ (i.e., the former cannot have a radius of convergence exceeding $1$). I leave it to you to show the radius of convergence equals ...



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