I'm looking for a way to get an estimate on a sum of the following series:
$$\sum_{i=1}^{n} \frac{1}{2i-1}$$
My exact question would be the solution for $n=500$ but I'd be interested in the generic solution as well.
$\endgroup$4 Answers
$\begingroup$Hint: Compare the sum $\sum_{i=1}^n \frac{1}{i}$ with the integrals $\int_1^{n} \frac1x \text{d}x$ and $\int_1^{n+1} \frac1x \text{d}x$.
Edit: The question has been changed from $\sum_{i=1}^{2n-1} \frac{1}{i}$ to $\sum_{i=1}^n \frac{1}{2i-1}$. Could you think of a similar trick here?
$\endgroup$ 2 $\begingroup$$$\sum_{i=1}^{n} \frac{1}{2i-1}=\frac{H_{n-\frac{1}{2}}}{2}+\log (2)$$ Now, expanding the harmonic number, for large values of $n$, the following approximation can be made $$\sum_{i=1}^{n} \frac{1}{2i-1}=\left(\frac{1}{2} \log (n)+\frac{\gamma }{2}+\log (2)\right)+\frac{1}{48 n^2}-\frac{7}{1920 n^4}+O\left(\left(\frac{1}{n}\right)^5\right)$$ Limited to the first term this gives, for $n=500$, an approximate value of $4.08905914555514$ while the exact value is $4.08905914555508\cdots$
$\endgroup$ 1 $\begingroup$If $s_n =\sum_{i=1}^{n} \frac{1}{2i-1} $, then
$\begin{array}\\ s_n &=\sum_{i=1}^{n} \frac{1}{2i-1}\\ &=\sum_{i=1}^{n} \left(\frac{1}{2i-1}+\frac{1}{2i}-\frac{1}{2i}\right)\\ &=\sum_{i=1}^{n} \left(\frac{1}{2i-1}+\frac{1}{2i}\right)-\sum_{i=1}^{n}\frac{1}{2i}\\ &=\sum_{i=1}^{2n} \frac{1}{i}-\frac12\sum_{i=1}^{n}\frac{1}{i}\\ &=H_{2n}-\frac12 H_n\\ \end{array} $,
where $H_n =\sum_{i=1}^{n}\frac{1}{i} \approx \ln n +\gamma+\frac1{2n}-\frac1{12n^2}+... $, so
$\begin{array}\\ s_n &\approx (\ln (2n) +\gamma+\frac1{4n}-\frac1{48n^2}+...) -\frac12(\ln n +\gamma+\frac1{2n}-\frac1{12n^2}+...)\\ &\approx (\ln n +\ln 2 +\gamma+\frac1{4n}-\frac1{48n^2}+...) -\frac12(\ln n +\gamma+\frac1{2n}-\frac1{12n^2}+...)\\ &\approx \frac12\ln n +\ln 2 +\frac12\gamma+\frac1{48n^2}+...\\ \end{array} $
$\endgroup$ 0 $\begingroup$Get the exact answer on Wolfram-Alpha by entering the description of the summation required as shown
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