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melese
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- Feb 24, 2012
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(HUN,1969) Prove that $\displaystyle1+\frac{1}{2^3}+\cdots+\frac{1}{n^3}<\frac{5}{4}$.
መለሰ
መለሰ
Let S=$\displaystyle1+\frac{1}{2^3}+\cdots+\frac{1}{n^3}+----$.(HUN,1969) Prove that $\displaystyle1+\frac{1}{2^3}+\cdots+\frac{1}{n^3}<\frac{5}{4}$.
መለሰ
Albert, thanks for your reply.Let S=$\displaystyle1+\frac{1}{2^3}+\cdots+\frac{1}{n^3}+----$.
S<$1+(\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+---+--)=1+\dfrac{1/8}{1-1/2}=\dfrac{5}{4}$
I did not assum tacitly $k^3\geq2^{k+1}$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 $k=2,3,4,5,6,7,8,9$!
መለሰ
That is what melese was telling you. Your inequality only works for the first few terms.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
melese and Opalg thanks both of you ,sorry ! I am so careless ,I did not check it ,I shouldThat is what melese was telling you. Your inequality only works for the first few terms.