Let $(a_n)$ be a sequence with the property that $\displaystyle \lim_{n \rightarrow \infty} n(a_{n+1}-a_n)=1$.
Prove that $\displaystyle \lim_{n \rightarrow \infty} a_n = \infty$ and $\sqrt[n]{a_n} = 1$
I just want to know if it's ok how I solved it.
Proof:
For the first part $\displaystyle \lim_{n \rightarrow \infty} n(a_{n+1}-a_n)=1 \Longrightarrow$ there exists a rank $n_\epsilon$ such that $\forall n>n_\epsilon$, $|n(a_{n+1}-a_n)-1|<\epsilon$, or $a_{n+1}-a_n>\frac{1-\epsilon}{n}$, equivalently $a_{n+1}>a_n+\frac{1}{n}-\frac{\epsilon}{n}$, $\forall n>n_\epsilon$. Taking $n$ to $\infty$ and summing up all the inequalities untill $n_\epsilon$,
we get $a_{n} > a_{n_\epsilon}$+ $(H_n-H_{n_\epsilon})-(n-n_\epsilon)\frac{\epsilon}{n}= a_{n_\epsilon}$+ $(H_n-H_{n_\epsilon})-(1-\frac{n_\epsilon}{n})\epsilon$. ($H_n$= is the n-th harmonic number). As $\displaystyle \lim_{n \rightarrow \infty}H_n = \infty$, is clear that $\displaystyle \lim_{n \rightarrow \infty} a_n =\infty$.
For the second part using Cauchy-D'alambert it's equivalent to finding the limit of $y_n=\frac{x_{n+1}}{x_n}$. The initial condition can be rewritten as $\displaystyle \lim_{n \rightarrow \infty} n a_n(y_{n}-1)=1$, so again there exists a rank $n_\epsilon$ such that $\forall n>n_\epsilon$, $|n a_n(y_{n}-1)-1|<\epsilon$ which means $(y_{n}-1)<\frac{1+\epsilon}{n a_n}$, but $\displaystyle \lim_{n \rightarrow \infty} n a_n= \infty$ so $(y_{n}-1)<\frac{1+\epsilon}{n a_n}<\epsilon$ which precisely means that $\displaystyle \lim_{n \rightarrow \infty} y_n=1$.
Are the approaches I used correct?
