I want to ask some problems that are asked in exam (actually the example of exam that I'm preparing in this post it is Mathematica analysis) I want some suggestions to my answer (some are obtained by reading other posts in here that I just aggregate)
I want to know if there are some suggestions and improvements and corrections. Thank you.
- Let $(a_n)$ be a positive real sequence such that $\lim_{n\rightarrow \infty}\frac{a_{n+2}}{a_n}=\frac{1}{2}$. Prove that $\lim_{n\rightarrow \infty} a_n=0$.
My first sol If $\lim_{n\rightarrow \infty}\frac{a_{n+2}}{a_n}=\frac{1}{2}$ we have $$\left|\dfrac{a_{n+2}}{a_n}-\dfrac{1}{2}\right|<\varepsilon,\forall n\ge N_0, \forall \varepsilon >0.$$ Let us choose $\varepsilon =1/4$ then for $n\ge N$ we have $\dfrac{a_{n+2}}{a_n}-\dfrac{1}{2}<\dfrac{1}{4}$, so $a_{n+2}<\dfrac{3}{4}a_n$ for all such $n$. Since $\lim_{m\rightarrow \infty}(3/4)^m=0$ (not sure if I could use it here withput proof), and by mathematical induction we have for all $k>N$ $$a_{2k}<\left(\dfrac{3}{4}\right)^{k-N}a_{2N},$$ $$a_{2k+1}<\left(\dfrac{3}{4}\right)^{k-N}a_{2N+1}.$$ Therefore, for all $n\ge 2N+M+1$, $a_n<\varepsilon$ as $(3/4)^{m}<\varepsilon/\max(a_{2N},a_{2N+1}), \forall \varepsilon>0, \forall m\ge M$ $\blacksquare$
My second sol By the ratio test we have both $\displaystyle \sum_{k\ge 1}a_{2k+1}$, $\displaystyle \sum_{k\ge 1}a_{2k}$ converges. If we let $b_k=a_{2k+1}$ and $c_k=a_{2k}$, then we have $b_k,c_k\rightarrow 0$ as $k\rightarrow \infty$. That is, $|a_{2k+1}|<\varepsilon$ and $|a_{2k}|<\varepsilon$ for all $\varepsilon$ and $n\ge N$, which means $|a_n|<\varepsilon$ $\blacksquare$
- Let $(a_n)$ be a positive real sequence such that $\sum_{n\ge 1} na_n<\infty$. Prove that $\sum_{m\le n}a_m $ is a Cauchy sequence.
I think it can be done by comparison test that $a_n\le na_n$ $\forall n\ge 1$, thus $\sum_{m\ge 1}a_m$ converges and so $\sum_{m\le n} a_m$ is a Cauchy sequence (not sure if this is okay since every convergent sequence is Cauchy.) This is something like if we let $s_n=\sum_{m\le n}a_m$, we have $\lim_{n\rightarrow\infty} s_n=c,\exists c$, and use the definiton again.
- Let $f:\mathbb R\rightarrow \mathbb R$ is continuous (does this mean it is continuous all over $\mathbb R?$) such that $f(0)>0$. Prove that $\exists \varepsilon>0$ and $\delta>0$ such that $f(x)>\varepsilon$ whenever $x$ satisfying $|x|<\delta$.
Since $f$ is continuous at $0$, $\forall\varepsilon,\exists \delta>0$, $|f(x)-f(0)|<\varepsilon$ whenever $|x|<\delta$. As $f(0)>0$ we let $f(0)=c,\exists c>0$, thus $f(x)>c-\varepsilon$. So we choose $\varepsilon=c/2$ so that $\exists\delta'>0$ we have $f(x)>c/2$ for any $|x|<\delta$ $\blacksquare$
- Let $f:\mathbb R\rightarrow\mathbb R$ is continuous at $a\in\mathbb R$. Let $g(x)=(x-a)f(x)$. Determine whether $g$ is differentiable at $a$ and explain why.
Since $f$ is continuous at $a$, we have that $|f(x)-f(a)|<\varepsilon$ for any $\varepsilon>0$ and some $\delta=\delta(\varepsilon)>0$ and for any $|x-a|<\delta$. For $h\neq 0$, we have $$\dfrac{g(a+h)-g(a)}{h}=\dfrac{hf(a+h)-0}{h}=f(a+h).$$ For any $|h|<\delta$, we have $|f(a+h)-f(a)|<\varepsilon$ so the limit above (as $h\to 0$) exists and equals $f(a)$ $\blacksquare$
let's first begin with your first solution: $\lim \frac{a_{n+2}}{a_{n}}=\frac{1}{2} \leftrightarrow \forall \epsilon >0,\exists N:n>N\implies \lvert \frac{a_{n+2}}{a_{n}}-\frac{1}{2} \rvert <\epsilon $ what u wrote before fixing $\epsilon $ is meaninless,u Start your proof by fixing $\epsilon=\frac{1}{4}$ ,this gives u $N$ ......,then u do the induction, after that u deduce(using the comparison test) that both $a_{2n}$ and $a_{2n+1}$ converge to zero,hence $a_{n}$ itself converges to zero.of course the choice of the rank $N$ depends on the value pf $\epsilon$,which is a bit messy in your arguments .the proof goes like this ,let $\delta >0$,then there are ranks $N,N"$ such that $n>N\implies \lvert a_{2n} \rvert <\delta \land n>N\implies \lvert a_{2n+1} \rvert <\delta$,Now let $s=\max\{N,N"\}$ ,then (Since any positive integer is either odd or even): $$\forall n>s \implies \lvert a_{n} \rvert <\delta $$,this proves that $\lim_{n\to \infty} a_{n}=0$. Regarding the second question, your anwser seems good .for the third question, choose directly $\epsilon=\frac{f(0)}{2}$,this is legitimate since $f(0)>0$ ,so $\exists \delta' \cdots$ and yes a function is continous iff it is continous at every point of its domain,for question 4,again u have a problem with quantifiers, $f$ continous at $a$ means $\forall \epsilon >0,\exists \delta ,\lvert x-a\rvert<\delta \implies \lvert f(x)-f(a)\rvert <\epsilon$ ,then the rest is right provided that u fix $\epsilon $ at the begining.