Thru the limit definition $(u_n)$ is the succession and $a$ is the limit of it :
$$ \left(u_n\right)\to\:a\:\:\Longleftrightarrow:\forall ε>0\:\:\:\exists p:\:\:\:\forall n\::\:\:\left(n>p\:\:\Rightarrow\:\left|u_n-a\right|< \:ε\right) $$
I need to prove the following:
$$ \frac{1}{\sqrt{n+1}}\rightarrow 0 $$
- Zero is my 'a', (to where the Succession is going) and $$ \frac{1}{\sqrt{n+1}}\ $$ is my '$\mathbf {u_n}$'
I don't even know from where should I start.
Thank you! :)
For any $\varepsilon >0$ we need to find a $p$ such that if $n>p$ then $\dfrac{1}{\sqrt{n+1}}<\varepsilon$
reverse the inequality
$\sqrt{n+1}>\dfrac{1}{\varepsilon}\to n+1>\dfrac{1}{\varepsilon^2}\to n>\dfrac{1}{\varepsilon^2}-1$
So if we take $p=\lceil \dfrac{1}{\varepsilon^2}-1\rceil$, for any $n>p$ it happens that $\dfrac{1}{\sqrt{n+1}}<\varepsilon$
For instance take $\varepsilon =10^{-6}$ we take $p=\lceil 10^{12} -1\rceil$ so if $n>p$ for instance if $n=10^{14}-1$ it happens that $\dfrac{1}{\sqrt{10^{14}}}=10^{-7}<10^{-6}$
Hope this helps