# Why does $\lim_{n\to\infty} \frac{a_{n+1}}{a_n} \neq 1$ if $(a_{n+1})$ is a subsequence of $(a_{n})$?

According to the quotient rule, if $\lim_{n\to\infty} \frac{a_{n+1}}{a_n} \lt 1$, then the sequence $(a_n)$ must be monotonically decreasing (definitively) and converge to $0$. If the limit is greater than $1$, then the sequence must be monotonically increasing and diverge.

I can't understand why the limit would ever be different than $1$, since if $a_n$ converges to $l$, then any subsequence of said sequence is also convergent and converges to the same limit $l$. Consequently, both $(a_n)$ and $(a_{n+1})$ converge to the same limit, and here lies the crux of my question: why can $\lim_{n\to\infty} \frac{a_{n+1}}{a_n} \neq 1$ if both sequences converge to the same limit $l$?

Please anyone shed some light on this as I really can't wrap my head around it.

• Both sequences converge to zero. – Lord Shark the Unknown Oct 29 '17 at 12:49
• Have you looked at any examples? Try $a_n=\frac 1{n!}$, say. – lulu Oct 29 '17 at 12:50
• I can't see how this would change anything. Isn't the limit of both $l$ anyway? – NTAuthority Oct 29 '17 at 12:51
• $\frac 00$ is not the same as $1$ – lulu Oct 29 '17 at 12:51
• @lulu Alright I see what you mean there – NTAuthority Oct 29 '17 at 12:57

Of course that $\lim_{n\to\infty} \frac{a_{n+1}}{a_n} \neq 1$ generally will happen.
The point is that $a_{n+1}$ and $a_n$ are both functions of $n$ but different ones and $a_{n+1}$ is "step ahead" of $a_n$ so they will take different values at the same $n$´s.
And, sometimes, this speed of convergence is fast enough to ensure $\lim_{n\to\infty} \frac{a_{n+1}}{a_n} \neq 1$.
For example, an example in comments is a nice one, we have $a_n=\frac 1{n!}$, but $a_{n+1}=\frac{1}{(n+1)!}=a_n \cdot \frac {1}{n+1}$.
Sometimes this speed will not be fast enough so we will sometimes have $\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = 1$