Question

1+ 1/3+ 1/9+....+1/3^(n-1) < 3/2??
Please, help

Answer 1

@ikram002p

Answer 2

1+ 1/3+ 1/9+....+1/3^(n-1) is a geometric series , u can easily find its sum , but whats ur question exactly ?

Answer 3

I have test: one of the problem is : if x1=1, \(x_{n+1}= 1+\dfrac{x_n}{3}\)
does \(X_n\) converge? if it is, find limit

Answer 4

Here was what I did
x2 = 4/3 > 1--> \(X_n\) increases for n=1,2
Assume X_n increases for n =k, k+1
need prove X_n increases for n = k+2

Answer 5

why don't you just use mathematical induction?

Answer 6

\(X_{k+1} > X_k\\\dfrac{X_{k+1}}{3}> \dfrac{X_k}{3}\\1+\dfrac{X_{k+1}}{3}> 1+\dfrac{X_k}{3}\)

Answer 7

the left hand side is \(X_{k+2}\) and the right hand side is \(X_{k+1}\)
that is \(X_n\) increases for all n in N

Answer 8

Now, claim \(X_n\) bounded by 3/2

Answer 9

\[1<\frac{3}{2} \text{ is true } \\ \text{ Suppose } 1+\frac{1}{3}+\frac{1}{9}+\cdots +\frac{1}{3^{n-1}}<\frac{3}{2} \text{ for integer } n \ge 0 \\ \text{ so we have } \\ 1+\frac{1}{3}+\cdots +\frac{1}{3^{n-1}}+\frac{1}{3^n} < \frac{3}{2}+\frac{1}{3^n} \\ \text{ but as } n->\infty , \frac{3}{2}+\frac{1}{3^n}->? \]

Answer 10

\(X_1= 1\\X_2= 1+ 1/3\\X_3= 1+1/3+1/9\\ ........\\X_n = 1+ 1/3+1/9+....+1/3^{n-1}\)

Answer 11

it leads to my question.

Answer 12

I just want to make sure that I have a right answer
since it is geometric series, so it sum = \(\dfrac{1-(1/3)^n}{1-(1/3)}\) < \(\dfrac{1}{1-(1/3)}= 3/2\)
and I am done.
Since \(X_n <3/2\) for all n, hence \(X_n\) bounded by 3/2

Answer 13

And finding out the limit is quite easy.

Answer 14

Tomorrow, I have to continue this test for 6 more questions. So that. I need confirm the stuff I was done.