MHB How can we prove the convergence of recursive defined sequences?

[mathmari]

Hey! :giggle:

a) Check the convergence of the sequence $a_n=\left (\frac{n+2000}{n-2000}\right)^n$, $n>1$. If it converges calculate the limit.
b) Check the convergence of the recursive defined sequence $a_n=\frac{a_{n-1}}{a_{n-1}+2}$, $n>1$, with $a_1=1$.For a) we have $$a_n=\left (1+\frac{4000}{n-2000}\right) ^{n-2000}\left (1+\frac{4000}{n-2000}\right) ^{2000}\to e^{4000}$$ Having found the limit means that the sequence is also convergent, right? but could we have shown the convergence also in an other way?For b) when we calculate some terms we see the sequence is decreasing and we can prove that using induction. It also holds that $a_n>0$. This means that the sequence converges, right?

:unsure:

 

[I like Serena]

a) Check the convergence of the sequence $a_n=\left (\frac{n+2000}{n-2000}\right)^n$, $n>1$. If it converges calculate the limit.

For a) we have $$a_n=\left (1+\frac{4000}{n-2000}\right) ^{n-2000}\left (1+\frac{4000}{n-2000}\right) ^{2000}\to e^{4000}$$ Having found the limit means that the sequence is also convergent, right? but could we have shown the convergence also in an other way?

Hey mathmari!

That works. (Nod)

We may want to mention which propositions we're using to conclude it though.
It's easy to make mistakes if we take the limits of parts of an expression after all.

I wouldn't immediately know a different way to do it, other than making sure the steps are correct.

b) Check the convergence of the recursive defined sequence $a_n=\frac{a_{n-1}}{a_{n-1}+2}$, $n>1$, with $a_1=1$.
For b) when we calculate some terms we see the sequence is decreasing and we can prove that using induction. It also holds that $a_n>0$. This means that the sequence converges, right?

Yep. It follows from the monotone convergence theorem. (Nod)

 

[mathmari]

That works. (Nod)

We may want to mention which propositions we're using to conclude it though.
It's easy to make mistakes if we take the limits of parts of an expression after all.

I wouldn't immediately know a different way to do it, other than making sure the steps are correct.Yep. It follows from the monotone convergence theorem. (Nod)

Great! Thank you! (Sun)