1
$\begingroup$

Prove that if $s_n$, $t_n$ are sequences such that $\lim\limits_{n \to \infty}s_n=+\infty$ and $t_n \ge 0$ for all n, then $\lim\limits_{n \to \infty}(s_n+t_n) = +\infty$.

My proof:

Since $\lim s_n=+\infty$, then for $M\,\gt0$, there exists $N$ such that $n\gt N\implies s_n\gt M$.

Now consider adding $t_n$ to each $s_n$ such that $s_n+t_n\gt M+t_n$ and since $t_n\ge 0$ for all $n$. For $t_n\gt 0$ this is true and for $t_n=0$ this is still true.

So for $\lim\,(s_n+t_n)$, let $A=M+t_n$ then for each $A$ there exists $N$ such that $n\gt N\implies s_n+t_n\gt A$ therefore $\lim\,(s_n+t_n)=+\infty$.

If this is wrong any hints or corrections would be very helpful.

$\endgroup$

1 Answer 1

2
$\begingroup$

There is a little issue with the fact that $A$ depends on $n$.

This is a more direct approach for the second half of your proof: for each $M>0$, let $N$ be as you defined ($n>N \implies s_n > M$). Then $n> N$ implies $s_n+t_n \ge s_n > M$.

$\endgroup$
3
  • $\begingroup$ What's the issue with A depending on n? $M+t_n$ is always positive right? since $t_n\ge 0$ $\endgroup$
    – reifi
    Nov 27, 2016 at 7:56
  • 1
    $\begingroup$ @reifi You need to show that for any constant $A$, that $s_n+t_n > A$ for all large $n$. If you choose $A$ to depend on $n$, you did not prove the previous statement. $\endgroup$
    – angryavian
    Nov 27, 2016 at 8:10
  • $\begingroup$ We can conclude $s_n + t_n > M + t_n $ but we can't conclude that $s_m+t_m > M+t_n $ for another value. But we do know $s_n + t_n \ge s_n > M$ for all n > N. We don't need a different A if M will do. $\endgroup$
    – fleablood
    Nov 28, 2016 at 4:10

You must log in to answer this question.

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