Proving Convergence: Solving the Limit of 1/(6n^2+1) = 0

[Mr Davis 97]

I am trying to show that $\displaystyle \lim \frac{1}{6n^2+1}=0$.

First, we have to find an N such that, given an $\epsilon > 0$, we have that $\frac{1}{6n^2+1} < \epsilon$. But in finding such an N, I get the inequality $n> \sqrt{\frac{1}{6}(\frac{1}{\epsilon}-1)}$. But clearly with the square root we would have to have that $\epsilon>1$, which doesn't make any sense since epsilon can be any positive real number. What am I doing wrong?

 

[FactChecker]

You got that wrong. To keep 1/ε -1 > 0, you need ε < 1.

 

[fresh_42]

You can be very generous here. You could e.g. take as well $\varepsilon^3$ and $n^3 > 6n^2 +1$. Maybe at the end there will be a maximum somewhere, as $2^3 \ngtr 6\cdot 2^2 +1$ and the starting point is say $N > 20$, but any estimates can be generous.

 

[Mr Davis 97]

You can be very generous here. You could e.g. take as well $\varepsilon^3$ and $n^3 > 6n^2 +1$. Maybe at the end there will be a maximum somewhere, as $2^3 \ngtr 6\cdot 2^2 +1$ and the starting point is say $N > 20$, but any estimates can be generous.

So I could use $\frac{1}{6n^2}$ just as well as $\frac{1}{6n^2+1}$ to prove convergence, since $\forall n \in \mathbb{N} \frac{1}{6n^2+1} < \frac{1}{6n^2}$?

 

[fresh_42]

You have to prove $\frac{1}{6n^2+1} < ... < \varepsilon$ for all $n > N=N(\varepsilon)$. The inequalities don't have to be close, only true. So $\frac{1}{6n^2+1} < \frac{1}{6n^2}$ as a first step is allowed.

 

[Mr Davis 97]

You have to prove $\frac{1}{6n^2+1} < ... < \varepsilon$ for all $n > N=N(\varepsilon)$. The inequalities don't have to be close, only true. So $\frac{1}{6n^2+1} < \frac{1}{6n^2}$ as a first step is allowed.

So in general, if $\forall n \in \mathbb{N} ~ a_n \ge b_n$ and $b_n$ converges, then $a_n$ converges?

 

[fresh_42]

So in general, if $\forall n \in \mathbb{N} ~ a_n \ge b_n$ and $b_n$ converges, then $a_n$ converges?

Why that? Simply take $a_n=n$ and $b_n=1$.

 

[FactChecker]

So in general, if $\forall n \in \mathbb{N} ~ a_n \ge b_n$ and $b_n$ converges, then $a_n$ converges?

You are being a little careless with your inequalities. You should say "and a_n converges, than bⁿ converges"

 

[fresh_42]

The idea is, that whatever one chooses for a radius $r$ of an open ball or interval around the limit, there will always be the rest of all sequence members within it, i.e. all up to finitely many (which are allowed to be outside). The members $a_1 ,\ldots , a_N$ are allowed outside, the members $a_{N+1},\ldots$ all must be inside. The only thing is, that this radius here is called $r=\varepsilon$. And of course the choice of $N$ depends on how small the ball / interval is; which is why I prefer to write $N(\varepsilon)$ instead. It reminds me on this fact.