I'm wondering if I'm solving the problems below correctly:
- Show that $$\sum_{k=1}^n \frac{1}{2k-1} = \ln (\sqrt{n}) + O(1)$$
Given $$\sum_{k=1}^n \frac{1}{k} = \ln(n) + O(1)$$ $$\sum_{k=1}^n \frac{1}{2k-1} + \sum_{k=1}^n \frac{1}{2k}- \sum_{k=1}^n \frac{1}{2k}$$ $$=\ln(2n) + O(1) - \frac{1}{2}\ln(n)-O(1)= \ln (\sqrt{n})+O(1)$$
- Evaluate $$\sum_{k=1}^\infty (2k+1)x^{2k}$$
$$\sum_{k=1}^\infty (2k+1)x^{2k} = \sum_{k=1}^\infty 2kx^{2k}+ \sum_{k=1}^\infty x^{2k}$$
Given $$\sum_{k=1}^\infty x^k=\frac{x}{1-x} \qquad \textrm{if } |x|<1$$ I think the second term is:
$$\sum_{k=1}^\infty x^{2k} = \frac{x^2}{1-x^2}$$
Differentiating the above on both sides and multiplying by $x$,
$$\sum_{k=1}^\infty 2kx^{2k} = \frac{2x^2}{(1-x^2)^2}$$
I added the two terms above and got $$\frac{3x^2-x^4}{(1-x^2)^2}$$
Did I compute this correctly?