Sequence problems - solution check

[Government$]

Homework Statement

Let $$(a_n)$$ be a sequence such that $$\lim_{n\rightarrow +\infty}n(a_n)=0$$.
1) What is
$$\lim_{n\rightarrow +\infty}(1 + {\frac{1}{n}} + (a_n))^n$$
2) For which value of p and l, after some n is $$(b_n)=\frac{n^{p \ cos(n\pi)}}{(1 + l + (a_n))^n}$$ properly defined. p and l are real numbers.

The Attempt at a Solution

1) Since $$\lim_{n\rightarrow +\infty}n(a_n)=0$$. I have that for some $$n_0$$, for every $$n>n_0$$ $$-1<n(a_n)$$ hence $$\lim_{n\rightarrow +\infty}\frac{n}{1 + n(a_n)} = \infty$$. Therefore i have that $$\lim_{n\rightarrow +\infty}(1 + {\frac{1}{n}} + (a_n))^n = \lim_{n\rightarrow +\infty}(1 + {\frac{1}{\frac{n}{1 + n(a_n)}}})^{\frac{n}{1 + n(a_n)}(1 + n(a_n))}= e^1=e$$

2) Considering that p is not diving anything here, and that p is exponent to n it seems to me that p can have any value. Now let $$l\neq -1 \neq 0$$. Since $$ (1 + l + (a_n)) = (1 + l + \frac{n(a_n)}{n})$$ and i have that $$ \lim_{n\rightarrow +\infty} \frac{n(a_n)}{n} = \lim_{n\rightarrow +\infty} \frac{(n+1)(a_{n+1}) - n(a_n)}{1} = \lim_{n\rightarrow +\infty} ((n+1)(a_{n+1}) - n(a_n))= 0 - 0 = 0 $$ by Stolz–Cesàro theorem. This means that I have for some $$n_0$$, for every $$n>n_0$$ So have that for $$1 + l > 0$$, $$|{\frac{n(a_n)}{n}}| < 1+l $$ and for $$1 + l < 0$$ $$|{\frac{n(a_n)}{n}}| < -(1+l) $$. Therefore $$(1 + l + \frac{n(a_n)}{n}) \neq 0$$ for $$n>n_0$$ so i won't have any divison by 0. For $$l=0$$ i also get $$(1 + \frac{n(a_n)}{n}) \neq 0$$ for $$n>n_0$$ since $$ \lim_{n\rightarrow +\infty} \frac{n(a_n)}{n} = 0$$. For $$l=-1$$ i get $$(b_n)=\frac{n^{n}n^{p \ cos(n\pi)}}{(n(a_n))^n}$$ Since $$\lim_{n\rightarrow +\infty}n(a_n)=0$$ i have no division by 0.
To conclude fore every p,l from R, after some $$n_0$$ for all $$n>n_0$$ $$(b_n)$$ is properly defined.

Sorry for any typos.

 

[STEMucator]

Homework Statement

Let $$(a_n)$$ be a sequence such that $$\lim_{n\rightarrow +\infty}n(a_n)=0$$.
1) What is
$$\lim_{n\rightarrow +\infty}(1 + {\frac{1}{n}} + (a_n))^n$$
2) For which value of p and l, after some n is $$(b_n)=\frac{n^{p \ cos(n\pi)}}{(1 + l + (a_n))^n}$$ properly defined. p and l are real numbers.

The Attempt at a Solution

1) Since $$\lim_{n\rightarrow +\infty}n(a_n)=0$$. I have that for some $$n_0$$, for every $$n>n_0$$ $$-1<n(a_n)$$ hence $$\lim_{n\rightarrow +\infty}\frac{n}{1 + n(a_n)} = \infty$$. Therefore i have that $$\lim_{n\rightarrow +\infty}(1 + {\frac{1}{n}} + (a_n))^n = \lim_{n\rightarrow +\infty}(1 + {\frac{1}{\frac{n}{1 + n(a_n)}}})^{\frac{n}{1 + n(a_n)}(1 + n(a_n))}= e^1=e$$

2) Considering that p is not diving anything here, and that p is exponent to n it seems to me that p can have any value. Now let $$l\neq -1 \neq 0$$. Since $$ (1 + l + (a_n)) = (1 + l + \frac{n(a_n)}{n})$$ and i have that $$ \lim_{n\rightarrow +\infty} \frac{n(a_n)}{n} = \lim_{n\rightarrow +\infty} \frac{(n+1)(a_{n+1}) - n(a_n)}{1} = \lim_{n\rightarrow +\infty} ((n+1)(a_{n+1}) - n(a_n))= 0 - 0 = 0 $$ by Stolz–Cesàro theorem. This means that I have for some $$n_0$$, for every $$n>n_0$$ So have that for $$1 + l > 0$$, $$|{\frac{n(a_n)}{n}}| < 1+l $$ and for $$1 + l < 0$$ $$|{\frac{n(a_n)}{n}}| < -(1+l) $$. Therefore $$(1 + l + \frac{n(a_n)}{n}) \neq 0$$ for $$n>n_0$$ so i won't have any divison by 0. For $$l=0$$ i also get $$(1 + \frac{n(a_n)}{n}) \neq 0$$ for $$n>n_0$$ since $$ \lim_{n\rightarrow +\infty} \frac{n(a_n)}{n} = 0$$. For $$l=-1$$ i get $$(b_n)=\frac{n^{n}n^{p \ cos(n\pi)}}{(n(a_n))^n}$$ Since $$\lim_{n\rightarrow +\infty}n(a_n)=0$$ i have no division by 0.
To conclude fore every p,l from R, after some $$n_0$$ for all $$n>n_0$$ $$(b_n)$$ is properly defined.

Sorry for any typos.

Your answer for part 1) is correct, namely this portion of it:

$$\lim_{n\rightarrow +\infty}(1 + {\frac{1}{n}} + (a_n))^n = \lim_{n\rightarrow +\infty}(1 + {\frac{1}{\frac{n}{1 + n(a_n)}}})^{\frac{n}{1 + n(a_n)}(1 + n(a_n))}= e^1=e$$

Though I'm not sure I agree entirely with the argument you used to arrive at this.

 

[Government$]

Though I'm not sure I agree entirely with the argument you used to arrive at this.

You mean the way i proved $$\lim_{n\rightarrow +\infty}\frac{n}{1 + n(a_n)} = \infty$$H
How about this:
$$\lim_{n\rightarrow +\infty}\frac{n}{1 + n(a_n)} = \lim_{n\rightarrow +\infty} n * \lim_{n\rightarrow +\infty} \frac{1}{1 + n(a_n)}=\infty*1=\infty$$. I have $$\lim_{n\rightarrow +\infty}\frac{n}{1 + n(a_n)} = \lim_{n\rightarrow +\infty} n * \frac{1}{1 + n(a_n)} = \lim_{n\rightarrow +\infty} n * \lim_{n\rightarrow +\infty} \frac{1}{1 + n(a_n)}$$ by limit properties.