Let $\left\{x_n\right\}$ be a sequence where $x_n = \frac{1}{1\cdot3}+\frac{1}{2\cdot5}+...\frac{1}{n\cdot(2n+1)}$
I have to calculate, to which point does the sequence $\left\{x_n\right\}$ converge, using Euler's constant($\gamma$). That is, I have to use the fact that the sequence $\left\{\gamma_n\right\}$ converges to $\gamma$, where $\gamma_n =1+\frac{1}{2}+...+\frac{1}{n}-\log n$.
But I cannot reduce $x_n$ to any form relating $\gamma_n$. Please help me solve it. Thanks in advance.
$$\begin{align} \sum_{k=1}^n\frac1{k(2k+1)} &=\sum_{k=1}^n\frac1{k}-2\sum_{k=1}^n\frac1{2k+1}\\ &=\sum_{k=1}^n\frac1{k}-2\left(\sum_{k=2}^{2n+1}\frac1k-\sum_{k=1}^n\frac1{2k}\right)\\ &=\sum_{k=1}^n\frac1{k}-2\sum_{k=2}^{2n+1}\frac1k+\sum_{k=1}^n\frac1k\\ &=2\sum_{k=1}^n\frac1{k}-2\left(-1+\sum_{k=1}^{2n+1}\frac1k\right)\\ &=2+2\sum_{k=1}^n\frac1{k}-2\sum_{k=1}^{2n+1}\frac1k\\ &=2+2(\gamma_n+\ln{(n)})-2(\gamma_{2n+1}+\ln{(2n+1)})\\ &=2+2\left(\gamma_n-\gamma_{2n+1}+\ln{\left(\frac{n}{2n+1}\right)}\right)\\ \end{align}$$ $$\begin{align} \therefore\lim_{n\to\infty}\sum_{k=1}^n\frac1{k(2k+1)} &=\lim_{n\to\infty}\left(2+2\left(\gamma_n-\gamma_{2n+1}+\ln{\left(\frac{n}{2n+1}\right)}\right)\right)\\ &=2+2\ln{\left(\frac12\right)}\\ &=\boxed{2(1-\ln{(2)})}\\ \end{align}$$