Let $a_{n} = \frac{1}{n^{2}}$ then

$ r = \lim_{n \to \infty} |\frac{a_{n}}{a_{n+1}}|= \lim_{n \to \infty}|\frac{(n+1)^{2}}{n^{2}}|$

Here why we behave to $\frac{1}{n^{2}}$ like as its a $\frac{n^{2}}{n^{2}}$ ?

Question

Let $a_{n} = \frac{1}{n^{2}}$ then

 $ r = \lim_{n \to \infty} |\frac{a_{n}}{a_{n+1}}|= \lim_{n \to \infty}|\frac{(n+1)^{2}}{n^{2}}|$

Here why we behave to $\frac{1}{n^{2}}$ like as its a $\frac{n^{2}}{n^{2}}$ ?

anonymous · Answer

in second lim...  why we have $$(n+1)^{2}$$ ?

anonymous · Answer

i think it should be $$1$$

anonymous · Answer

actually i know when we see a n we should do it n+1  but where is our 1 from $$\frac{1}{n^{2}}$$

kinggeorge · Answer

You wrote: $$r = \lim_{n \to \infty} \left|\;\frac{a_{n}}{a_{n+1}}\right|$$It should be$$r = \lim_{n \to \infty} \left|\;\frac{a_{n+1}}{a_{n}}\right|$$

anonymous · Answer

hmm in original question its like i wrote, maybe our mentor write it wrong... but my question is how we can use quotient rule to $$\frac{1}{n^{2}}$$

anonymous · Answer

they get above $$\frac{(n+1)^{2}}{n^{2}}$$ but how its possible ? :(

kinggeorge · Answer

I've got leave after this, but if you use $$r = \lim_{n \to \infty} \left|\;\frac{a_{n+1}}{a_{n}}\right|$$you can get to the following (after some simplification).$$r=\lim_{n \to \infty} \left|\;\frac{n^2}{(n+1)^2}\right|=\lim_{n \to \infty} \;\frac{n^2}{(n+1)^2}$$Use l'hopitals rule on this, and you get a solution for $r$.

anonymous · Answer

ok thanks George :)