"Let $\alpha ,\beta \in (0,1),\alpha +\beta =1,\{a_{n}\}$ and $\ \{b_{n}\}$ real, nonnegative, bounded sequences. If \begin{equation*} \lim_{n\rightarrow \infty }[\alpha a_{n}+\beta b_{n}]\text{ exist } \end{equation*}

then $\lim_{n\rightarrow \infty }a_{n}$ and $\lim_{n\rightarrow \infty }b_{n}$ exist?"

cannot prove it or cannot give counterexample please help me

  • $\begingroup$ How are $\alpha$ and $\beta$ quantified? Is it for all $\alpha$ or $\beta$ or for some $\alpha$ and $\beta$? $\endgroup$ – Michael Burr May 18 '17 at 19:30
  • $\begingroup$ I think $\alpha$ and $\beta$ are supposed to be constants...? $\endgroup$ – Frpzzd May 18 '17 at 19:32
  • $\begingroup$ yes they are constant $\endgroup$ – seniormath May 18 '17 at 19:34
  • $\begingroup$ $$\frac12\cos^2n+\frac12\sin^2n$$ $\endgroup$ – Yves Daoust May 18 '17 at 19:42
  • $\begingroup$ Given divergent sequence a and alpha=beta=.5 choose b=-a . $\endgroup$ – Jacob Wakem May 18 '17 at 20:00

As a simple counterexample, consider some $0\le b_n=1-a_n\le1$ known to have no limit vs. $\frac12a_n+\frac12b_n$. (Could be $a_n=\sin^2n$.)

  • $\begingroup$ "$\alpha,\beta \in (0,1)$" $\endgroup$ – adfriedman May 18 '17 at 19:35
  • $\begingroup$ @gt6989b Fixed. $\endgroup$ – Yves Daoust May 18 '17 at 19:37
  • $\begingroup$ upvoted already $\endgroup$ – gt6989b May 18 '17 at 19:38

Consider the following counterexample:

Let $\alpha , \beta = 0.5 $.
Let $(a_n)_n$ be such that $a_n = 1$ for even $n$ and $a_n = 0$ for odd $n$.
Let $(b_n)_n$ be such that $b_n = 1$ for odd $n$ and $b_n$ = 0 for even $n$.

Clearly $\lim_{n\to \infty}(\alpha a_n + \beta b_n)_n = 1$ for it is a constant sequence.
Since $(a_n)_n$ and $(b_n)_n$ are alternating sequences their limits do not exist.


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