# Prove that $\sum_{k=1}^{n}{\frac{4k}{(2k-1)(2k+1)}}-\ln(n)\rightarrow \ln(4)+\gamma-1$

Let $\gamma$ denote Euler's constant. Prove that $\sum_{k=1}^{n}{\frac{4k}{(2k-1)(2k+1)}}-\ln(n)\rightarrow \ln(4)+\gamma-1$.

I don't know whether I can somehow use result from $\sum_{k=1}^{n}{\frac{1}{2k-1}}-\frac{1}{2}\ln(n)\rightarrow\ln(2)+\frac{\gamma}{2}$ in this problem. Or I should try to use telescopic series? Can anyone give me some hints?

• $\frac{4k}{(2k-1)(2k+1)} = \frac{1}{2k+1}+\frac{1}{2k-1}$ – sharding4 Jun 27 '17 at 19:19
• $$\begin{eqnarray*}\sum_{k=1}^{n}\left(\frac{1}{2k-1}+\frac{1}{2k+1}\right)&=&H_{2n-1}+H_{2n+1}-\frac{H_{n-1}}{2}-\frac{H_n}{2}-1\\&=&2 H_{2n}-H_n+\frac{1}{2n+1}-1\tag{1} \end{eqnarray*}$$ and since $H_m = \log(m)+\gamma+o(1)$ the asymptotic behaviour of $(1)$ is given by $$o(1)+2\log(2)+\log(n)+\gamma-1\tag{2}$$ what is difficult? – Jack D'Aurizio Jun 27 '17 at 19:40

Well, $$\frac{4k}{(2k-1)(2k+1)}=\frac1{2k-1}+\frac1{2k+1}.$$ You know the asymptotic for $$\sum_{k=1}^n\frac1{2k-1}.$$ But the series $$\sum_{k=1}^n\frac1{2k+1}$$ is almost the same as the previous one, so its asymptotic is easily deduced.
Partial fractions gives \begin{eqnarray*} \frac{4k}{(2k-1)(2k+1)}=\frac{1}{2k-1}+\frac{1}{2k+1} \end{eqnarray*} Note that this gives all the odd recriprocals twice (apart the first one) ... also add & subtract the even terms (twice) \begin{eqnarray*} -1 +\underbrace{\sum_{i=1}^{n} \frac{2}{2i-1} -\color{red}{\sum_{i=1}^{n} \frac{2}{2i}}}_{\underbrace{2\sum_{i=1}^{n} \frac{(-1)^{i-1}}{i}}_{2 \ln2}}+\underbrace{\color{red}{\sum_{i=1}^{n} \frac{1}{i}}- \ln(n)}_{\gamma} \end{eqnarray*} Now the first two sums can be combined & will give in the limit $2 \ln2$ and the third sum less $ln n$ will give (in the limit) $\gamma$ (The Euler-Maschroni constant).