# ${{a}_{n}}=\frac{1}{2n-1}$，${{S}_{n}}=\sum\limits_{i=1}^{n}{{{a}_{i}}}$，if ${{S}_{n}}<3$，Calculate max(n).

${{a}_{n}}=\frac{1}{2n-1}$, ${{S}_{n}}=\sum\limits_{i=0}^{n}{{{a}_{i}}}$, if ${{S}_{n}}<3$, Calculate max(n).

sorry.n is from 1, not 0

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Thank you for your answer.A student asked.I estimate should be without a computer is not complete. – tianzhidaosunyouyu Sep 25 '12 at 5:48

The following may not be accurate enough, but it will give a good estimate. Let $H_m$ the the $m$-th harmonic number, that is, $$H_m=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{m}.$$

It is known that for $m$ large (and it doesn't have to be very large), we have $H_m\approx \log m+\gamma$, where $\gamma$ is the Euler-Mascheroni constant, roughly $0.5772$.

Note that $S_n=1+\frac{1}{3}+\cdots+\frac{1}{2n-1}$. Add in the missing even denominator terms, up to and including $\frac{1}{2n}$. These add up to $\frac{1}{2}H_n$. It follows that $$S_n=H_{2n}-\frac{1}{2}H_n \approx (\log 2n+\gamma)-\frac{1}{2}(\log n+\gamma).$$ But $\log 2n=\log 2+\log n$. Thus $$S_n\approx \frac{1}{2}\log n +\frac{1}{2}\gamma+\log 2.$$ If we want $S_n\approx 3$, we want $\log n\approx 6-\gamma-2\log 2.$ That gives the estimate $n\approx 56.6$. If this estimate is good enough, the largest $n$ should be $56$.

There is information available on the the error in the approximation $H_m\approx \log m+\gamma$ that would probably enable a definite answer.

Much simpler direct calculation settles the problem in a simple way, but I wanted to describe the harmonic number approach.

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looks like it is 56

sum from $k=1..n ( 1/(2*k - 1) ) = 3, 1<n<10000$

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$S_{56} <3$, $S_{57}>3$.

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