Estimated value of $\sum \limits_{i=1}^{50} \frac{1}{2i-1}$? Anyone has any idea the estimated result(no need to be accurate) of $\sum \limits_{i=1}^{50} \frac{1}{2i-1}$? I think this problem is not hard but I don't have a generalized method to solve!
And I also remember this problem related to a very famous probability problem, any one has any idea? Thank you!
 A: You can generalize this method:
Consider the recurrence relation:
$$f(x+1)=f(x)+\frac{1}{2x-1}$$
$$f(0)=1$$
The first equality can be written as:
$$f(x+1)-f(x)=\frac{1}{2x-1}$$
Then if we sum both sides from $x=0$ to $n-1$ notice we have:
$$\sum_{x=0}^{n-1} (f(x+1)-f(x))=\sum_{x=0}^{n-1} \frac{1}{2x-1}$$
Using the idea of a telescoping sum and $f(0)$ we have for $n-1 \geq 1 \implies n \geq 2$:
$$f(n)=1+\sum_{x=0}^{n-1} \frac{1}{2x-1}=\sum_{x=1}^{n-1} \frac{1}{2x-1}$$
But considering the antiderivative of $\frac{1}{2n-1}$  for $n \geq 1$ and the definition of the derivative we have:
$$\frac{\ln (2(n+1)-1)}{2}-\frac{\ln (2n-1)}{2} \approx  \frac{1}{2n-1}=f(n+1)-f(n)$$
Hence:
$$f(n) \approx \frac{\ln (2n-1)}{2}+C$$
$$f(n+1)=\sum_{x=1}^{n} \frac{1}{2x-1} \approx \frac{\ln (2(n+1)-1)}{2}+C$$
Where now the issue becomes finding the best constant $C$. 
A good thing to do would be to let $C$ give you the best approximation as $n \to \infty$:
$$C:=\lim_{n \to \infty} \left((\sum_{x=1}^{n} \frac{1}{2x-1})-\frac{\ln (2n+1)}{2} \right)$$
Through numerical computation I believe:
$$C=.6351....$$
Therefore:
$$\sum_{x=1}^{n} \frac{1}{2x-1} \approx \frac{\ln (2n+1)}{2}+.6351..$$
So we get the approximate value:
$$\approx \frac{\ln (2(50)+1)}{2}+.6351=\frac{\ln (101)}{2}+.6351 \approx 2.94266..$$
Which is close to the actual value of:
$$2.93777...$$
(Wolfram alpha evaluates $C$ to be equal to $\frac{1}{2}(\ln (2)+\gamma)$ where $\gamma$ denotes the Euler-Mascheroni constant, which is defined in a similar way we defined $C$)
A: Using BCLC's approach and answer, this can be generalized to $$\sum_{i=1}^n\frac 1{2i-1}=H_{2n}-\frac 12 H_n$$ For large values of $n$, we can use the asymptotics of the harmonic numbers $$H_m=\gamma +\log(m) +\frac{1}{2 m}-\frac{1}{12
   m^2}+O\left(\frac{1}{m^4}\right)$$ which them makes $$\sum_{i=1}^n\frac 1{2i-1}=H_{2n}-\frac 12 H_n=\left(\frac{\gamma }{2}+\log
   (2)\right)+\frac{1}{2} \log (n)+\frac{1}{48 n^2}+O\left(\frac{1}{n^4}\right)$$ where $\gamma$ denotes the Euler-Mascheroni constant.
For $n=50$, the exact value would be $$\frac{3200355699626285671281379375916142064964}{10893808629642574556958407646142547
   43075}\approx 2.937774848475$$ while the approximating formula would give $$\left(\frac{\gamma }{2}+\log
   (2)\right)+\frac{1}{2} \log (50)+\frac{1}{120000}\approx 2.937774849058$$
A: Observe that
$$\frac11 + \frac13 + \frac15 + \cdots + \frac1{2(50)-1}$$
$$ = \left( \frac11 + \frac12 + \frac13 + \frac14 + \cdots + \frac1{2(50)} \right) - \left( \frac12 + \frac14 + \cdots + \frac1{2(50)} \right)$$
$$ = \left( \frac11 + \frac12 + \frac13 + \frac14 + \cdots + \frac1{2(50)} \right) - \frac12\left( \frac11 + \frac12 + \cdots + \frac1{50} \right) $$
$$= H_{100} - \frac12H_{50}$$
where $H_n$ is the nth Harmonic number:
$$H_n = \sum_{i=1}^{n} \frac{1}{i}$$
