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- Apr 14, 2013

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Hey!!

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?

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?

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