Prove A < B without Calculation Tools

[Albert1]

A=$1001^{999}$

B=$1000^{1000}$

Prove :A<B

(note :any calculation tools are not allowed ,also no use of log function)

 

[MarkFL]

Re: prove A<B

I would use the binomial theorem to write:

$$1001^{999}=(1000+1)^{999}=\sum_{k=0}^{999}{999 \choose k}1000^{999-k}$$

Since we have 1000 terms, and each term would have to be equal to $1000^{999}$ in order for the summation to be equal to $1000^{1000}$, yet for $0<k$ we find:

$${999 \choose k}<1000^k$$

we may therefore conclude that:

$$1001^{999}<1000^{1000}$$

 

[Opalg]

Re: prove A<B

A=$1001^{999}$

B=$1000^{1000}$

Prove :A<B

(note :any calculation tools are not allowed ,also no use of log function)

Are we allowed to use the fact that $\bigl(1+\frac1n\bigr)^n$ increases to $e$ as $n\to\infty$? If so, then $\bigl(1+\frac1n\bigr)^n < n+1$ whenever $n\geqslant2$. It follows that $(n+1)^n < (n+1)n^n$ and therefore $(n+1)^{n-1} < n^n.$

 

[mathworker]

Re: prove A&lt;B

I would use the binomial theorem to write:

$$1001^{999}=(1000+1)^{999}=\sum_{k=0}^{999}{999 \choose k}1000^{999-k}$$

Since we have 1000 terms, and each term would have to be equal to $1000^{999}$ in order for the summation to be equal to $1000^{1000}$, yet for $0<k$ we find:

$${999 \choose k}<1000^k$$

we may therefore conclude that:

$$1001^{999}<1000^{1000}$$

as you are using $$1000^k$$under sigma i think you can't compare $${999 \choose k}$$ with
$$1000^k$$...

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is it correct to use $$1000^K$$ under sigma

 

[Albert1]

Re: prove A<B

I like Opalg 's method (my solution is similar to his)

 

[MarkFL]

Re: prove A&lt;B

as you are using $$1000^k$$under sigma i think you can't compare $${999 \choose k}$$ with
$$1000^k$$...

- - - Updated - - -

is it correct to use $$1000^K$$ under sigma

My intended purpose was to recognize that for $0<k$, we have:

$${999 \choose k}1000^{999-k}<1000^{999}$$

and since:

$$\sum_{k=0}^{999}1000^{999}=1000\cdot1000^{999}=1000^{1000}$$

we must therefore have:

$$\sum_{k=0}^{999}{999 \choose k}1000^{999-k}=1001^{999}<1000^{1000}$$

 

[mathworker]

Re: prove A&lt;B

My intended purpose was to recognize that for $0<k$, we have:

$${999 \choose k}1000^{999-k}<1000^{999}$$

and since:

$$\sum_{k=0}^{999}1000^{999}=1000\cdot1000^{999}=1000^{1000}$$

we must therefore have:

$$\sum_{k=0}^{999}{999 \choose k}1000^{999-k}=1001^{999}<1000^{1000}$$

okay i got it, you are comparing each term to $$1000^999$$...THANK YOU