1.

a) Given the sequence $ { \dfrac{n^2+1}{2n^2-1}}$ prove that there exists an order $ { k}$ where $ { \left | u_n - \dfrac{1}{2} \right |<10^{-3}}$ is valid.


$ { \begin{aligned} \left | \dfrac{n^2+1}{2n^2-1} - \dfrac{1}{2} \right | &< 10^{-3} \\ \left | \dfrac{2n^2+2-2n^2+1}{2(2n^2-1)} \right | &< 10^{-3} \\ \left | \dfrac{3}{2(2n^2-1)} \right | &< 10^{-3} \\ \dfrac{|3|}{|2(2n^2-1)|} &< 10^{-3} \end{aligned}}$


Since $ { 2(2n^2-1)>0}$ what follows is


$ { \begin{aligned} \dfrac{3}{2(2n^2-1)} &< 10^{-3} \\ 3/2 \times 10^3 &< 2n^2-1 \\ 3/4\times 10^3+1/2 &< n^2 \\ \sqrt{3/4\times 10^3 + 1/2} &< n \end{aligned}}$


So by taking $ { k > \left \lfloor \sqrt{3/4\times 10^3 + 1/2}\right \rfloor +1}$ we have the intended result.


b) Prove by definition that $ { u_n \rightarrow 1/2}$


By the definition of limit, and using a), we have $ { n > \sqrt{\dfrac{3}{4 \delta}+1/2}}$. If we take $ { k= \left \lfloor \sqrt{\dfrac{3}{4 \delta}+1/2} \right\rfloor+1}$ the difference between $ { u_n}$ and $ { 1/2}$ is always less than $ { \delta}$.


2. Prove that $ { \lim u_n = 0 \Leftrightarrow \lim |u_n| = 0}$


The gist of this result is that sometimes is easier to prove that the modulus of a sequence tends to zero than to prove the sequence tends to zero. Since one result implies the other we may avoid doing some boring calculations.


We say that $ { u_n \rightarrow a}$ iff $ { \forall \delta > 0 \, \exists k \in \mathbb{N}: \quad n>k \Rightarrow |u_n - a| < \delta}$


Thus $ { \lim |u_n - a| = 0}$ iff

$ {\forall \delta > 0\,\exists k\in\mathbb{N}:\; n > k\Rightarrow||u_n-a|-0| < \delta}$


$ {\Leftrightarrow \forall \delta > 0 \, \exists k \in \mathbb{N}:\; n > k \Rightarrow |u_n - a| < \delta}$


Hence with $ { a=0}$ the conditions $ { \lim u_n = 0}$ and $ { \lim |u_n| = 0}$ are indeed equivalent.


3. Calculate $ { \lim \sqrt{n+1}-\sqrt{n}}$


This limit we are interested in calculating can be seen as $ { \lim u_n - v_n}$ where $ { u_n = \sqrt{n+1}}$ and $ { v_n = \sqrt{n}}$. We know that $ { \lim u_n = \lim \sqrt{n+1} = +\infty}$ and $ { \lim v_n = \lim \sqrt{n} = +\infty}$.


So what we are trying to measure is how fast these sequences diverge. If this limit is $ { a \in \mathbb{R}^+}$ then $ { u_n}$ grows slightly faster, if it is $ { a \in \mathbb{R}^-}$ then it is $ { v_n}$ that grows slightly faster.


In the case of $ { \pm \infty}$ we have that $ { u_n}$, for the $ { +}$ sign ($ { v_n}$ for the $ { -}$ sign) tends infinitely faster to the given limit.


On with the calculations now:


$ {\begin{aligned} \lim \sqrt{n+1}-\sqrt{n} &= \lim \dfrac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}} \\ &= \lim \dfrac{n+1-n}{\sqrt{n+1} + \sqrt{n}} \\ &= \lim \dfrac{1}{\sqrt{n+1}+\sqrt{n}} \\ &= \lim \dfrac{1}{\sqrt{n(1+1/n)}+\sqrt{n}} \\ &= \lim \dfrac{1}{\sqrt{n}\sqrt{1+1/n}\sqrt{n}} \\ &= \lim \dfrac{1}{\sqrt{n}\left( \sqrt{1+1/n}+1 \right)} \\ &= \lim\dfrac{1}{\left( \sqrt{1+1/n}+1 \right) } \lim \dfrac{1}{\sqrt{n}} \\ &= \lim\dfrac{1}{2 \sqrt{n}} \\ &= 0 \end{aligned}}$


4. Calculate $ { \lim \left( \sqrt{n^2+n} - \sqrt{n^2+1} \right)}$


$ {\begin{aligned} \lim \left( \sqrt{n^2+n} - \sqrt{n^2+1} \right)&=\lim \dfrac{n^2+n-n^2-1}{\sqrt{n^2+n} + \sqrt{n^2+1}} \\ &=\lim \dfrac{n-1}{\sqrt{n^2\left(1+\frac{1}{n}\right)} + \sqrt{n^2\left(1+\frac{1}{n^2}\right)}} \\ &=\lim \dfrac{n-1}{n \sqrt{1+\frac{1}{n}} + n\sqrt{1+\frac{1}{n^2}}} \\ &=\lim \dfrac{n-1}{n\left( \sqrt{1+\frac{1}{n}} + \sqrt{1+\frac{1}{n^2}} \right)} \\ &=\lim \dfrac{n-1}{2n} \\ &=\dfrac{1}{2} \end{aligned}}$


5. Calculate $ { \lim \displaystyle \sum _{k=1}^n \dfrac{1}{(n+k)^2}}$.


Let us write out a few terms of this sum so that we can gain some intuition of what's going on


$ { \displaystyle \sum _{k=1}^n \dfrac{1}{(n+k)^2} = \dfrac{1}{(n+1)^2}+\dfrac{1}{(n+2)^2}+\cdots + \dfrac{1}{(2n)^2}}$


So by making $ { n \rightarrow \infty}$ what we get is a bigger number of increasingly small numbers to add.


The result of this limit will tell us what of these two contradictory effects wins.


Since we are adding up $ { n}$ terms that get increasingly small we have


$ \displaystyle \displaystyle \sum _{k=1}^n \dfrac{1}{(n+k)^2} \leq \dfrac{n}{(n+1)^2}$

And also



$ \displaystyle \displaystyle \sum _{k=1}^n \dfrac{1}{(n+k)^2} \geq \dfrac{n}{4n^2}$

Hence $ { \dfrac{n}{4n^2} \leq \displaystyle \sum _{k=1}^n \dfrac{1}{(n+k)^2} \leq \dfrac{n}{(n+1)^2}}$ with $ { \lim \dfrac{n}{4n^2} = \lim \dfrac{n}{(n+1)^2} = 0}$


Thus $ { \lim \displaystyle \sum _{k=1}^n \dfrac{1}{(n+k)^2} = 0}$ also.


Hence the fact that the fractions tend to $ { \,0}$ is more relevant to the value of the limit than the the fact that we get an infinite number of fractions to add.


6. Calculate $ { \displaystyle \sum _{k=1}^n \dfrac{1}{\sqrt{n+k}}}$


Now we have a similar situation to the previous exercise but this time the numbers that are being added are bigger than the previous ones. So it begs the question: What will be the limit this time?

Like previously we are summing $ { n}$ increasingly small terms and so



$ \displaystyle \displaystyle \sum _{k=1}^n \dfrac{1}{\sqrt{n+k}} \geq \dfrac{n}{\sqrt{2n}}$

And



$ \displaystyle \displaystyle \sum _{k=1}^n \dfrac{1}{\sqrt{n+k}} \leq \dfrac{n}{\sqrt{n+1}}$

Thus $ { \dfrac{n}{\sqrt{2n}} \leq \displaystyle \sum _{k=1}^n \dfrac{1}{\sqrt{n+k}} \leq \dfrac{n}{\sqrt{n+1}}}$.


Since $ { \lim \dfrac{n}{\sqrt{2n}} = \lim \dfrac{1}{\sqrt{2}}\dfrac{n}{\sqrt{n}}= \lim \dfrac{1}{\sqrt{2}}\sqrt{n} = + \infty}$ and $ { \lim \dfrac{n}{\sqrt{n+1}} = \lim \dfrac{n}{\sqrt{n}}\dfrac{1}{\sqrt{1+1/n}} = \lim \sqrt{n}\dfrac{1}{\sqrt{1+1/n}} = +\infty}$ it follows that $ { \displaystyle \sum _{k=1}^n \dfrac{1}{\sqrt{n+k}} = + \infty}$


This time the fact that the number of fractions to add grows without a bound is more relevant than the fact that those numbers tend to $ { \,0}$.


7. Calculate $ { \lim \dfrac{n^n}{n!}}$

Let us do this visually:


$ { n^{n-1} = n \times n \times n \ldots \times n}$ with $ { n-1}$ terms.


$ { n! = 1 \times 2 \times 3 \times \ldots \times n = 2 \times 3 \times \ldots \times n}$ with $ { n-1}$ terms.


So $ { \lim \dfrac{n^n}{n!} \geq \lim \dfrac{n^n}{n^{n-1}} = + \infty}$. Then also $ { \lim \dfrac{n^n}{n!} = +\infty}$


From this we can see that $ { n^n}$ grows to infinity faster than $ { n!}$


8. Give examples of sequences that

a) $ { u_n \rightarrow +\infty}$ and $ { v_n \rightarrow - \infty}$: $ { u_n+v_n=0}$

$ { u_n = n}$ and $ { v_n = -n}$


b) $ { u_n \rightarrow +\infty}$ and $ { v_n \rightarrow - \infty}$: $ { u_n+v_n=10}$

$ { u_n = n+10}$ and $ { v_n = -n}$


c) $ { u_n \rightarrow +\infty}$ and $ { v_n \rightarrow - \infty}$: $ { u_n+v_n=+\infty}$

$ { u_n = 2n}$ and $ { v_n = -n}$


d) $ { u_n \rightarrow +\infty}$ and $ { v_n \rightarrow - \infty}$: $ { u_n+v_n}$ doesn't exist.

$ { u_n = n+(-1)^n}$ and $ { v_n = -n}$


e) $ { u_n \rightarrow 0}$ and $ { v_n \rightarrow \infty}$: $ { u_n v_n = a \in \mathbb{R}}$

$ { u_n = \dfrac{a}{n}}$ and $ { v_n = n}$


f) $ { u_n \rightarrow 0}$ and $ { v_n \rightarrow \infty}$: $ { u_n v_n = 0}$

$ { u_n = \dfrac{1}{n^2}}$ and $ { v_n = n}$


g) $ { u_n \rightarrow 0}$ and $ { v_n \rightarrow \infty}$: $ { u_n v_n = +\infty}$

$ { u_n = \dfrac{1}{n}}$ and $ { v_n = n^2}$


h) $ { u_n \rightarrow 0}$ and $ { v_n \rightarrow \infty}$: $ { u_n v_n}$ doesn't exist

$ { u_n = \dfrac{\sin n}{n}}$ and $ { v_n = n}$