# Showing this sequence converges.

Given a sequence $x_n=\left(\dfrac{2n^3+n}{n^3} \right)+ i\left(\dfrac{3n}{n+1}\right)$, how would I show it converges? How would I choose $N$?

I did the following.

Given $\epsilon >0,$ choose $N>[?]$. Then for $n>N$

\begin{align} \left \lvert x_n-(2+3i) \right \rvert & = \left \lvert \frac{2n^3+n}{n^3}-2 \right \rvert+ \lvert \frac{3n}{n+1}-3 \rvert\\ & = \left \lvert \frac{1}{n^2} \right \rvert+ \left \lvert -\frac{3}{n+1} \right \rvert\\ & =\frac{1}{n^2}+\frac{3}{n+1} \\ & = \frac{n+1+3n^2}{n^2+n^3} <?<\epsilon \end{align}

Hence $x_n \rightarrow 2+3i$.

-
You will need to work with the usual distance between complex numbers, $|(a_n+ib_n)-(2+3i)|$ is the square root of $(a_n-2)^2+(b_n-3)^2$. – André Nicolas May 19 '12 at 18:12
This is basically correct. The first $=$ should be a $\leq$, and for the last ? use that $1, n < n^2$ for the numerator and $n^2 + n^3 > n^3$ for the denominator. It should be clear now how to pick $N$ as a function of $\epsilon$. – Zarrax May 19 '12 at 18:13
Thanks guys. @AndréNicolas , thanks, does that mean I have to square the whole thing whenever I deal with sequences involving complex numbers? – Steven May 20 '12 at 1:19
In p4rinciple, in this case, you have to show that for large enough $n$, $\sqrt{(a_n-2)^2 +(b_n-3)^2}<\epsilon$. And for an early exercise that's exactly what you are expected to do. Later you can take shortcuts. – André Nicolas May 20 '12 at 1:25
Okay, thanks andre! – Steven May 20 '12 at 16:16

The sequence, as you have guessed right, converges to $2+3i$. Your argument is also almost fine. You need to clarify which norm you are using to prove convergence. Usually one uses the Euclidean norm, in which case, you get \begin{align} \left \lVert x_n - (2+3i) \right \rVert_2 & = \left \lVert \left( \frac{2n^3+n}{n^3} + \frac{3n}{n+1} i \right) - (2+3i) \right \rVert_2\\ & = \left \lVert \left( \frac{2n^3+n-2n^3}{n^3} \right) + i \left( \frac{3n-3(n+1)}{n+1} \right)\right \rVert_2\\ & = \left \lVert \left( \frac{1}{n^2} \right) - i \left( \frac{3}{n+1} \right)\right \rVert_2\\ & = \sqrt{\left( \frac1{n^2}\right)^2 + \left(\frac{3}{n+1} \right)^2} \end{align} Note that for $n \in \mathbb{Z}^+$, we have $\dfrac1{n^2} < \dfrac3{n+1}$. Hence, we get that $$\sqrt{\left( \frac1{n^2}\right)^2 + \left(\frac{3}{n+1} \right)^2} < \sqrt{\left(\frac{3}{n+1} \right)^2 + \left(\frac{3}{n+1} \right)^2} = \frac{3 \sqrt{2}}{n+1}$$ Now given an $\epsilon > 0$, choose $N(\epsilon) = \dfrac{3\sqrt{2}}{\epsilon} - 1$. Now for all $n \geq N$, where $n \in \mathbb{Z}^+$, we have that $$\dfrac{3 \sqrt{2}}{n+1} < \epsilon.$$ Hence, given an $\epsilon > 0$, choose $N(\epsilon) = \dfrac{3\sqrt{2}}{\epsilon} - 1$. Now for all $n \geq N$, where $n \in \mathbb{Z}^+$, we have that $$\lVert x_n - \left( 2+3i\right)\rVert_2 < \epsilon$$ Hence, $x_n \rightarrow 2+3i$.

EDIT

You could also use the $1$-norm as you have. \begin{align} \left \lVert x_n - (2+3i) \right \rVert_1 & = \left \lVert \left( \frac{2n^3+n}{n^3} + \frac{3n}{n+1} i \right) - (2+3i) \right \rVert_1\\ & = \left \lVert \left( \frac{2n^3+n-2n^3}{n^3} \right) + i \left( \frac{3n-3(n+1)}{n+1} \right)\right \rVert_1\\ & = \left \lVert \left( \frac{1}{n^2} \right) - i \left( \frac{3}{n+1} \right)\right \rVert_1\\ & = \left \lvert \left( \frac1{n^2} \right)\right \rvert + \left \lvert \left(\frac{3}{n+1} \right) \right \rvert \end{align} Note that for $n \in \mathbb{Z}^+$, we have $\dfrac1{n^2} < \dfrac3{n+1}$. Hence, we get that $$\left \lvert \left( \frac1{n^2} \right)\right \rvert + \left \lvert \left(\frac{3}{n+1} \right) \right \rvert < \left \lvert \left( \frac{3}{n+1} \right)\right \rvert + \left \lvert \left(\frac{3}{n+1} \right) \right \rvert = \frac{6}{n+1}$$ Now given an $\epsilon > 0$, choose $N(\epsilon) = \dfrac{6}{\epsilon} - 1$. Now for all $n \geq N$, where $n \in \mathbb{Z}^+$, we have that $$\dfrac{6}{n+1} < \epsilon.$$ Hence, given an $\epsilon > 0$, choose $N(\epsilon) = \dfrac{6}{\epsilon} - 1$. Now for all $n \geq N$, where $n \in \mathbb{Z}^+$, we have that $$\lVert x_n - \left( 2+3i\right)\rVert_1 < \epsilon$$ Hence, $x_n \rightarrow 2+3i$.

-
@ Marvis , how would be the convergence with respect to $L^p$ norms ? – Theorem May 19 '12 at 18:59
@Ananda It will be the same. Replace $2$ by $p$ and the same argument will carry you through. Probably I should just add it to the post instead of repeating the argument for $2$ and $1$. – user17762 May 19 '12 at 19:27
Thanks @Marvis, most appreciated! I think I was meant to work in the 1-norm, but thanks for the heads up for the 2 norm! May I ask, your choice of $N$ is different from donantonio (below), but generally when we find trying to choose an $N$, is there a 'fixed'/'unique' answer? Or as long as what we choose holds its okay? – Steven May 20 '12 at 1:16

You reached the point where you want to prove $\,\displaystyle{\frac{3n^2+n+1}{n^3+n^2}<\epsilon}\,$ , but $$\frac{3n^2+n+1}{n^3+n^2}\leq \frac{5n^2}{n^3}\leq \frac{5}{n}$$ so $\,\displaystyle{\frac{5}{n}<\epsilon \Longrightarrow n>\frac{5}{\epsilon}}\,$ , and thus it is enough to choose $\displaystyle{\,N_\epsilon:=\left[\frac{5}{\epsilon}\right] + 1}\,$ to have that

for any $\,n>N_\epsilon\,$ we have the wanted inequality (with [x] = the integer part of x)

-
Thanks @DonAntonio , may I ask your choice of $N$ is different from Marvis's answer, so when we find trying to choose an $N$, is there a 'fixed'/'unique' choice for $N$? Or as long as what we choose for $N$ holds/is valid then it is okay? – Steven May 20 '12 at 1:18
Well, obviously when you've found one such N, any other choice greater than this one will work as the definition of limit requires the existence of "some" such N, and there's no mathematical point, imo, to ask "the minimal one", or "the best one" (other than, perhaps, to make students sweat and/or force them to practice their basic algebra skills). As you saw in the different answers, it is usual to make evaluations of the expression, some finer than others, so the results can widely differ. – DonAntonio May 20 '12 at 1:46
Thanks for clearing things up, understood now. – Steven May 20 '12 at 16:16

For sequences of complex numbers it occurs that a given sequence of complex numbers $a_{n}=x_{n}+y_{n}i$ converges to some complex number $x+yi$ iff sequence $x_{n}$ converges to $x$ and as one may expect $y_{n}$ converges to $y$. Sequence diverges if at least one of the mentioned conditions does not hold.

-
Thanks, appreciated. :) – Steven May 20 '12 at 1:19