suppose we have sequences $(a_n)$, $b_n=a_{n+1}$, and $(c_n)$ where $c_n\neq 0 \forall n$. Furthermore $(c_n)$ has the property where $\lim_{n\to\infty}(c_n/c_{n+1})=1$.

One can show that $\lim_{n\to\infty}a_n = L \iff \lim_{n\to\infty}b_n = L$.

Let $x= Lc$ where $L$ is the limit of $(a_n)$ and $(b_n)$, and $c$ is the limit of $(c_n)$. Then

$\lim_{n\to\infty}a_nc_n = Lc =x \iff \lim_{n\to\infty}b_nc_n = Lc =x $

therefore if $(b_nc_n)$ converges, then $(a_nc_n)$ converges to the same limit

However, I'm not sure how it is possible to find an example where $(a_nc_n)$ and $(b_nc_n)$ converges to different values, since $a_n$ and $b_n$ have the same limit, then shouldn't $(a_nc_n)$ and $(b_nc_n)$ converge to the same value no matter what by the algebraic limit theorem?

  • 1
    $\begingroup$ It is not given that $(c_n) $ is convergent. $\endgroup$ Apr 24, 2021 at 7:24
  • $\begingroup$ @KaviRamaMurthy How can one prove that $(a_kc_k)$ converges then? If we consider $a_k(c_k/c_{k+1})=L\cdot 1$ by the algebraic limit theorem, then $a_kc_k=Lc_{k+1}$. However, this doesn't yield anything useful. $\endgroup$ Apr 24, 2021 at 7:29
  • $\begingroup$ Are you sure it is $(c_n/c_{n+1}) \to 1$ and not $|c_n/c_{n+1}| \to 1$ ? $\endgroup$
    – nicomezi
    Apr 24, 2021 at 7:32
  • $\begingroup$ @nicomezi yes, its given that $(c_n/c_{n+1})\to 1$ and not $|c_n/c_{n+1}|\to 1$ $\endgroup$ Apr 24, 2021 at 7:34

1 Answer 1


Your argument is not correct since you assumed convergence of $(c_n)$ However, your guess is right. $(a_nc_n)$ and $(b_nc_n)$ cannot converge to different limits:

$b_nc_n=a_{n+1}c_n=\frac {a_{n+1}c_{n+1}} {c_{n+1}/c_n} $ which shows that $(a_nc_n)$ and $(b_nc_n)$ cannot have different limits. If either of these sequences converges so does the other, with the same limit.

  • $\begingroup$ I see, this also implies that it is not possible for $(a_nc_n)$ to converge while $(b_nc_n)$ diverges, correct? $\endgroup$ Apr 24, 2021 at 7:42
  • $\begingroup$ @bruhmoment28 Yes, if one of then converges the other one also converges and the limits are same. $\endgroup$ Apr 24, 2021 at 7:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .