Inequality

[melese]

(HUN,1969) Prove that $\displaystyle1+\frac{1}{2^3}+\cdots+\frac{1}{n^3}<\frac{5}{4}$.

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[Albert]

Re: inequality

(HUN,1969) Prove that $\displaystyle1+\frac{1}{2^3}+\cdots+\frac{1}{n^3}<\frac{5}{4}$.

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Let S=$\displaystyle1+\frac{1}{2^3}+\cdots+\frac{1}{n^3}+----$.
S<$1+(\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+---+--)=1+\dfrac{1/8}{1-1/2}=\dfrac{5}{4}$

 

[melese]

Re: inequality

Let S=$\displaystyle1+\frac{1}{2^3}+\cdots+\frac{1}{n^3}+----$.
S<$1+(\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+---+--)=1+\dfrac{1/8}{1-1/2}=\dfrac{5}{4}$

Albert, thanks for your reply.

I think you've assumed tacitly $k^3\geq2^{k+1}$ for $k\geq2$, but it's only true for $k=2,3,4,5,6,7,8,9$!

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[Albert]

Re: inequality

Albert, thanks for your reply.

I think you've assumed tacitly $k^3\geq2^{k+1}$ for $k\geq2$, but it's only true for $k=2,3,4,5,6,7,8,9$!

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I did not assum tacitly $k^3\geq2^{k+1}$

I only constructed a geometric series with ratio 1/2 (1 not included) and

compare terms by terms

ex:1=1 , 1/8 =1/8

1/27 <1/16

1/64<1/32

1/125<1/64

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and the infinite geometric series (first term=1/8 , and ratio =1/2) sums up to 1/4

so S<1+1/4=5/4

 

[Opalg]

Re: inequality

I did not assum tacitly $k^3\geq2^{k+1}$

I only constructed a geometric series with ratio 1/2 (1 not included) and

compare terms by terms

ex:1=1 , 1/8 =1/8

1/27 <1/16

1/64<1/32

1/125<1/64

1/216 < 1/128

1/343 < 1/256

1/512 = 1/512 Things are not looking so good already!

1/729 < 1/1024 NOT TRUE!!

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and the infinite geometric series (first term=1/8 , and ratio =1/2) sums up to 1/4

so S<1+1/4=5/4

That is what melese was telling you. Your inequality only works for the first few terms.

 

[Albert]

Re: inequality

That is what melese was telling you. Your inequality only works for the first few terms.

melese and Opalg thanks both of you ,sorry ! I am so careless ,I did not check it ,I should

have checked it before ,do you have a better solution for this inequality ?

May be we should use "fourier series" to prove it

 

[Opalg]

Re: inequality

Using partial fractions, $\dfrac1{n^3} < \dfrac1{n(n-1)(n-2)} = \dfrac{1/2}n - \dfrac1{n-1} + \dfrac{1/2}{n-2}.$

Apply that for $n = 4,\ 5,\ 6, \ldots$, to get $$\begin{aligned} S &= 1 + \frac1{2^3} + \frac1{3^3} + \frac1{4^3} + \ldots \\ &< 1 + \frac18 +\frac1{27} + \Bigl(\frac{1/2}4 - \frac13 + \frac{1/2}2\Bigr) + \Bigl(\frac{1/2}5 - \frac14 + \frac{1/2}3\Bigr) + \Bigl(\frac{1/2}6 - \frac15 + \frac{1/2}4\Bigr) + \ldots \\ &= 1 + \frac18 +\frac1{27} + \frac14 - \frac16\quad \text{ (all the subsequent terms cancel)} \\ &< 1 + \frac18 +\frac1{24} + \frac14 - \frac16 = \frac54. \end{aligned}$$

 

[melese]

Re: inequality

We could prove a stronger inequality, namely $\displaystyle1+\frac{1}{2^3}+\cdots+\frac{1}{n^3}<\frac{5}{4}-\frac{1}{4n}$.
Proceeding by induction on $n$. Clearly it's true when $n=1$.
Soppose it's true for some $n$. Then, $\displaystyle1+\frac{1}{2^3}+\cdots+\frac{1}{n^3}+\frac{1}{(n+1)^3}<\frac{5}{4}-\frac{1}{4n}+\frac{1}{(n+1)^3}=\frac{5}{4}-(\frac{1}{4n}-\frac{1}{(n+1)^3})$. Hence, to show that $\displaystyle1+\frac{1}{2^3}+\cdots+\frac{1}{n^3}+\frac{1}{(n+1)^3}<\frac{5}{4}-\frac{1}{4(n+1)}$, it's enuogh to show that $\displaystyle\frac{1}{4n}-\frac{1}{(n+1)^3}\geq\frac{1}{4(n+1)}$, but that's a simple matter of exercise....

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