What is the sum of this fraction series? $$\frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{9} + \frac{1}{11} .... + \frac{1}{99}?$$ Is there a way to do without calculation? I would appreciate an answer that gives an easy way to find this sum.
Define $$H_k=\frac{1}{1}+\frac{1}{2}+\cdots+\frac{1}{k}$$

These are called harmonic numbers.  Then your sum is $H_{100}-\frac{1}{2}H_{50}-1$.

If you're happy with an approximation, we know that $H_k\approx \ln k + \frac{1}{2k}+\gamma$, where $\gamma\approx  0.5772156$.  Hence your answer is approximately $\ln 100 + \gamma-0.5 \ln 50 -\frac{\gamma}{2}-1\approx 1.93776648$

You can get an exact answer on Wolfram Alpha, but it isn't particularly nice.  The above approximation is correct to four decimal places.
You could probably use simple sequence addition then simplify afterwords,
For example $1+3+5......99$ is equivalent to $((99+1)/2)^2$ which is $2500$ so the denominator would be $2500$ then find the number of terms and add the numerator.