Show that $a_n = \frac{n}{2n+1}+\frac{1}{n^3}$ is a Cauchy Sequence

Question

I want to show that

$$a_n = \frac{n}{2n+1} + \frac{1}{n^3}$$

is a Cauchy sequence.

My attempt:

$$|a_m-a_n|=|(\frac{m}{2m+1}-\frac{n}{2n+1}+\frac{1}{m^3}-\frac{1}{n^3})|$$ $$\leq\frac{m}{2m+1}+\frac{n}{2n+1}+\frac{1}{m^3}+\frac{1}{n^3}$$ $$<\frac{1}{2}+\frac{1}{2}+\frac{1}{N^3}+\frac{1}{N^3}$$ $$=1+\frac{2}{N^3}$$ $$\leq \epsilon\Rightarrow N(\epsilon)\geq(\frac{2}{\epsilon-1})^{\frac{1}{3}}$$

But this seems a bit off?

Any thoughts?

Answer 1

You got it wrong here:

$$\begin{align}|a_m - a_n| &= \left|\frac{m}{2m+1}-\frac{n}{2n+1}+\frac{1}{m^3}-\frac{1}{n^3}\right| \\ &\leqslant \frac{m}{2m+1}+\frac{n}{2n+1}+\frac{1}{m^3}+\frac{1}{n^3} \end{align}$$

I mean, it is correct, but it won't help. The number inside the absolute value bars is small precisely because of those subtractions, and expanding it the way you did makes those subtractions vanish.

You should try to keep the subtractions around:

$$\begin{align}|a_m - a_n| &= \left|\frac{m}{2m+1}-\frac{n}{2n+1}+\frac{1}{m^3}-\frac{1}{n^3}\right| \\ &\leqslant \left|\frac{m}{2m+1}-\frac{n}{2n+1}\right|+\left|\frac{1}{m^3}-\frac{1}{n^3}\right| \end{align}$$

Can you go on from here? [Hint: fix $m\leqslant n$. The other case, $n<m$, is similar.]

Answer 2

You split too much, instead:

\begin{align}\left|\frac{m}{2m+1}-\frac{n}{2n+1} + \frac{1}{m^3}-\frac{1}{n^3}\right| &\leq \left|\frac{m}{2m+1}-\frac{n}{2n+1}\right| + \left|\frac{1}{m^3}-\frac{1}{n^3}\right|\\ &\leq \left|\frac{m(2n+1) - n(2m+1)}{(2m+1)(2n+1)}\right| + \left|\frac{n^3-m^3}{m^3n^3}\right|\\ &= \frac{|m-n|}{(2m+1)(2n+1)} + \frac{|m-n||m^2+mn+n^2|}{m^3n^3} \end{align}

I'll let you take it from here...