My Math Forum Determining if sequences converge or diverge

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 April 9th, 2013, 06:21 AM #1 Member   Joined: Mar 2013 Posts: 56 Thanks: 0 Determining if sequences converge or diverge Hallo! please help me with these, i am struggling and they are for revision purposes determine is sequences {avn} (n is subscript) converge or diverge an justify answers 1) an = (((-1)^n) + n)/(((-1)^n) - n) 2) an = sqrt(n+1) - sqrt(n) thank you in advance!
 April 9th, 2013, 08:58 AM #2 Senior Member   Joined: Feb 2009 Posts: 172 Thanks: 5 Re: Determining if sequences converge or diverge They both converge. (a) $a_n=\frac{(-1)^n+n}{(-1)^n-n}=\frac{n$$\frac{(-1)^n}{n}+1$$}{n$$\frac{(-1)^n}{n}-1$$}=\frac{$$\frac{(-1)^n}{n}+1$$}{$$\frac{(-1)^n}{n}-1$$}$. Since $\displaysty\lim_{n\rightarrow\infty}\frac{(-1)^n}{n}=0$ we have that $\displaysty\lim_{n\rightarrow\infty}\frac{(-1)^n}{n}+1=1$ and $\displaysty\lim_{n\rightarrow\infty}\frac{(-1)^n}{n}-1=-1$. Hence, $\displaysty\lim_{n\rightarrow\infty}a_n=\displayst y\lim_{n\rightarrow\infty}\frac{$$\frac{(-1)^n}{n}+1$$}{$$\frac{(-1)^n}{n}-1$$}=-1$. (b) $a_n=\sqrt{n+1}-\sqrt{n}=$$\sqrt{n+1}-\sqrt{n}$$\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\s qrt{n}}=\frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{ n}}$. Since $\displaystyle\lim_{n\rightarrow\infty}\sqrt{n+1}=+ infty$ e $\displaystyle\lim_{n\rightarrow\infty}\sqrt{n}=+in fty$ we have that $\displaystyle\lim_{n\rightarrow\infty}\sqrt{n+1}+\ sqrt{n}=+infty$ wich gives us that $\displaystyle\lim_{n\rightarrow\infty}a_n=\display style\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n+1}+ \sqrt{n}}=0$.
 April 9th, 2013, 03:24 PM #3 Member   Joined: Mar 2013 Posts: 56 Thanks: 0 Re: Determining if sequences converge or diverge Excellent thank you so much! I was very stuck on this

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### a subscript n 1=sqrtbn

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