${{a}_{n}}=\frac{1}{2n-1}$，${{S}_{n}}=\sum\limits_{i=1}^{n}{{{a}_{i}}}$，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
 A: 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. 
A: looks like it is 56
sum from $k=1..n  ( 1/(2*k - 1) ) = 3, 1<n<10000$
A: $S_{56} <3$, $S_{57}>3$.
