Question

Is it converge ? Also calculate ..

* the question in attachments

Answer 1

@nincompoop can you help me , please ?

Answer 2

I think it diverges for \(|m|\ge1\)

and converges for \(|m| \lt 1\)

Answer 3

Can you tell me how you got to this result?

Answer 4

\[\begin{align}
a_{n+1}&=\color{blue}{a_n}+m^n\\~\\
&=\color{blue}{a_{n-1}+m^{n-1}}+m^n\\~\\
&=a_{n-2}+m^{n-2}+m^{n-1}+m^n\\~\\
&\vdots\\~\\
&=a_0 + m^0+m^1+m^2+\cdots+m^{n-2}+m^{n-1}+m^n\\~\\
&=a_0+\frac{m^{n+1}-1}{m-1}

\end{align}\]

Answer 5

\[\begin{align}
a_{n+1}&=a_0+\frac{m^{n+1}-1}{m-1}\\~\\
\end{align}\]

thats the solution to given recurrence relation, take the limit as \(n\to\infty\) and see for what values of \(m\) the sequence converges

Answer 6

the sequence is
a0 = 1
a1 = 1 + m^0 = 1 + 1
a2 = 1 + 1 + m^1
a3 = 1 + 1 + m^1 + m^2
...
an+1 = 1 + 1+ m + m^2 + ... + m^n

this is a geometric sequence
the sum is 1 + 1 / ( 1- m) , and it converges for |m| <1 , diverges for |m| >= 1

Answer 7

@perl @rational .. Thank you very much :))

can you help me of this , please ?

Answer 8

It is a fibonacci sequence
are you trying to find an explicit formula ?

Answer 9

what is the ( fibonacci sequence ) ?

Answer 10

it is a special sequence that starts with 1,1 and the next term can by obtained by adding previous two terms

Answer 11

it doesn't converge as you can see the terms keep growing

Answer 12

aha , can we find ( calculate ) the limit of it ?

Answer 13

there wont be a limit as it diverges

Answer 14

sometimes you could say the limit is \(+\infty\)

Answer 15

yup it diverges

Answer 16

Ok , Thank you all very much :)

Answer 17

you are welcome :)