# Convergence of a sequence problem

Assume that the sequence $a_n$ converges to $l$. Let $b_n = a_{n+1}/n$. Show that $b_n$ converges to the same $l$.

Since $a_n$ converges to $l$, for any $\varepsilon> 0$, there is a natural $N$ such that if $n ≥ N$, then $|a_n − l| < \varepsilon$. Note that if $n ≥ N$, then $n+1≥ N$ and $|a_{n+1} − l| < \varepsilon$ as well. That is, if $n ≥ N$, we have $|b_n − l| < \varepsilon$. This proves that $b_n$ converges to l as well.

I found this proof in a book but I modified $b_n$ from $b_n =a_{n+1}$ to $b_n = a_{n+1}/n$ and now I am stuck. What do I do about the $1/n$ the $m$.

• Please use LaTeX for readability. – Hanul Jeon Feb 8 '18 at 5:35
• I don't understand why you modified the definition of $b_n$. With the modified definition $b_n \to 0$, not to $l$. – Kavi Rama Murthy Feb 8 '18 at 6:01
• The result is totally false. It should be obvious why. – zhw. Feb 8 '18 at 23:11

I don't quite understand your question. Actually, if $a_n$ converges to a real number $l$, and $b_n = a_n/n$, here is a proof that $$\lim_{n\to\infty}b_n = 0$$
Solution 1. We have $\lim_{n\to\infty}a_n = l$ and $\lim_{n\to\infty}\frac 1 n=0$. Use the multiplication rule, we have $$\lim_{n\to\infty}b_n = \lim_{n\to\infty}a_{n+1} \cdot \lim_{n\to\infty}\frac 1 n= l \cdot 0= 0$$
Solution 2. $\forall\epsilon>0$. Since $a_n$ converges to $l$, there exists $N_0\in\mathbb{N}$ such that $\forall n>N_0$, $|a_n-l|<l$. Let $N=\max\{\lceil\frac{2l}{\epsilon}\rceil,N_0\}$. Thus, for all natural number $n>N$ (which implies $n+1>n>N\ge N_0$) $$0<a_{n+1}<2l$$ multiply by $1/n$ to both side of the inequality, $$0<\frac{a_{n+1}}{n}=b_n<\frac{2l}{n}<\frac{2l}{N}\le\epsilon$$ Therefore, $\forall\epsilon>0\exists N\in\mathbb{N}\forall n\in\mathbb{N} (n>N\Rightarrow|b_n-0|<\epsilon)$, which means $\lim_{n\to\infty}b_n = 0$.