real analysis

**flower3** · September 11th 2009, 08:38 AM

${ For n \in N \ , let x_n = 1+ \frac{1}{2^3} +\frac{1}{3^3} + ...+\frac{1}{n^3} , and \ let \ n_k = 2^k - 1 , k \in N }$

$a- prove \ that \ x_{n_k} < \ 1 + (\frac {1}{4})+(\frac {1}{4})^2+...+(\frac {1}{4})^{k-1} , k=2,3,4,...$

$b- use \ (a ) \ to \ prove \ that \ x_n \ is \ convergent$

**Taluivren** · September 11th 2009, 03:59 PM

Hi,
we can prove a) by induction:
base case: $k=2$ .... $n_2=3$ .... $x_{n_2}=1+\frac{1}{2^3}+\frac{1}{3^3}<1+\frac{1}{8 }+\frac{1}{8} = 1+\frac{1}{4}$

inductive step: suppose $x_{n_k} = 1+ \frac{1}{2^3} +\frac{1}{3^3} + ...+\frac{1}{(2^k-1)^3}< 1 + \frac {1}{4}+\frac {1}{4^2}+...+\frac {1}{4^{k-1}}=y_k$ and let's prove the statement holds for $k+1$ :

$x_{n_{k+1}}= 1+ \frac{1}{2^3} +\frac{1}{3^3} + ...+\frac{1}{(2^{k+1}-1)^3} =$
$=\underbrace{1+ \frac{1}{2^3} +\frac{1}{3^3} + ...+\frac{1}{(2^k-1)^3}}_{<y_k}+\frac{1}{(2^k)^3}+\frac{1}{(2^k+1)^3 }+\frac{1}{(2^k+2)^3}+...+\frac{1}{(2^{k+1}-1)^3}<$

$<y_k +\underbrace{\frac{1}{(2^k)^3}}_{=\frac{1}{2^{3k}} }+\underbrace{\frac{1}{(2^k+1)^3}}_{<\frac{1}{2^{3 k}}}+\underbrace{\frac{1}{(2^k+2)^3}}_{<\frac{1}{2 ^{3k}}}+...+\underbrace{\frac{1}{(2^{k+1}-1)^3}}_{<\frac{1}{2^{3k}}}<$

$<y_k+ \left( (2^{k+1}-1)-2^k+1 \right)\frac{1}{2^{3k}} = y_k+(2\cdot 2^k-2^k)\frac{1}{2^{3k}} = y_k+\frac{2^k}{2^{3k}}=$

$= y_k+\frac{1}{2^{2k}}= y_k+\frac{1}{4^k}= 1 + \frac {1}{4}+\frac {1}{4^2}+...+\frac {1}{4^{k-1}}+\frac{1}{4^k} = y_{k+1}$

so the inductive step is completed and a) holds.

As for b), from a) we see that that the sequence $x_{n_k}$ is bounded from above by $y_k$ .
$y_k$ is the sum of first $k$ terms of a geometric progression, $y_k$ converges to 4/3.
So since $x_n$ is increasing, it is convergent.

**flower3** · September 12th 2009, 08:29 AM

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