Question

I have a convergent sequence \[\left\{ \frac{ 3n+4i n }{ n+3 } \right\}, n=0,1,2...infinity \]
And I need to find the limit value a.
Furthermore, then I need to find the lowest value\[N \in \mathbb{N} \] which meets the requirements:
\[\left| \frac{ 3n+4i n }{ n+3 }-a \right|\le \frac{ 1 }{ 1000 } \forall n \ge N\]

Answer 1

I can simply input it to find the limit: \[\lim_{n \rightarrow infinity}a_n=a\] But is that enough of a "proof" / I feel like i need to show it more than just using CAS?

Answer 2

\(\large a_n= \frac{ 3n+4 }{ n+3 } \)

is that the expression ?

Answer 3

No, \[a_n=\frac{ 3n+4i n }{ n+3 }\]

Answer 4

Oh, \(i\) is the imaginary unit is it ?

Answer 5

Yes

Answer 6

\[a_{n}=3+4i-\frac{ 9+12i }{ n+3}\]

lim_{n}->infty =3+4i
can we do this?

Answer 7

that looks good to me!

Answer 8

Yeap, that seems like a smart ideer. How did you get to that?

Answer 9

It's like you have to manipulate with the sequences into something smart, that you can work with. And I generally find that part the hardest.

Answer 10

well i thought to remove n from the numerator

so made a term like this- \[\frac{ 1 }{ n+3}\]we know that the numerator should have 3n+4in

so we have to add something that gives us 3n+4in in the numerator :)

so now comes the twisty part
we add 3+4i
we get this-\[(3+4i) +\frac{ 1 }{ n+3 }\]
simplifying we will be getting this-\[\frac{ 3n+9+4i n+12i }{n+3 }\]
but here we see that 9 +12i is the waste part that we don't need

so we need to modify our term a little bit such that the 9+12i gets canceled out so we do this-\[\frac{ -(9+12i) }{ 3+n }\]

now when we add 3+4i the results will be-\[(3+4i)+\frac{ -(9+12i) }{ n+3 }\]\[\frac{ 3n+4i n+9+12i-9-12i }{ n+3}=>\frac{ 3n+4i n}{n+3 }\]

Answer 11

this way we get the term \[(3+4i)-\frac{ (9+12i) }{ n+3 }\]

Answer 12

Ohh okay, thanks.
So with the limit, we need to isolate n in the equation:
\[\left| \frac{ 3n+4i n }{ n+3 }-(3+4i) \right|\le \frac{ 1 }{ 1000 }\]
Is there a clever way to start with such inequality?

Answer 13

\[3n+4in=3(n+3-3)+4i(n+3-3)=(3+4i)(n+3)-(9+12i)\]

Answer 14

Yea, that makes abit more sense for me @Zarkon thanks. But adding and subtracting 3, is that just a "great idea" or how did you come up with that?

Answer 15

I think that the distance is given by the module of the complex number

Answer 16

so we can write this:
\[\left| {\frac{{3n + 4in}}{{n + 3}} - (3 + 4i)} \right| = \frac{{15}}{{n + 3}} \leqslant \frac{1}{{1000}}\]

Answer 17

Did you first solve: \[\frac{ 3n+4i n }{ n+3 }\]
Using the conjugated denominator? @Michele_Laino

Answer 18

I assume you found modulus with the formula: \[\left| x+iy \right|=\sqrt{x^2+y^2}\] But is this possible, when there is 2 complex numbers inside?

Answer 19

denominator is a real number, so I applied this identity:
\[\left| {\frac{z}{a}} \right| = \frac{1}{{\left| a \right|}}\left| z \right|,\quad a \in \mathbb{R}\]
where:
\({\left| a \right|}\) stands for absolute value of \(a\)
whereas \(|z|\) stands for the module of \(z\)

Answer 20

So you found modulus of \[\frac{ 3n+4i n }{ n+3 }\] and \[-(3+4i)\] separately?

Answer 21

no, please I have computed the value of the fraction inside the absolute value, here are my steps:
\[\begin{gathered}
\left| {\frac{{3n + 4in}}{{n + 3}} - (3 + 4i)} \right| = \left| {\frac{{\left( {3n + 4in} \right) - (3 + 4i)\left( {n + 3} \right)}}{{n + 3}}} \right| = \hfill \\
\hfill \\
= \left| {\frac{{3n + 4in - 3n - 9 - 4in - 12i}}{{n + 3}}} \right| = \hfill \\
\hfill \\
= \left| {\frac{{ - 9 - 12i}}{{n + 3}}} \right| = \frac{{15}}{{n + 3}} \hfill \\
\end{gathered} \]

Answer 22

Ohh okay, thank you very much.

Answer 23

:)