1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ $\frac{4k}{(2k-1)(2k+1)} = \frac{1}{2k+1}+\frac{1}{2k-1}$ $\endgroup$ – sharding4 Jun 27 '17 at 19:19
  • 1
    $\begingroup$ $$\begin{eqnarray*}\sum_{k=1}^{n}\left(\frac{1}{2k-1}+\frac{1}{2k+1}\right)&=&H_{2n-1}+H_{2n+1}-\frac{H_{n-1}}{2}-\frac{H_n}{2}-1\\&=&2 H_{2n}-H_n+\frac{1}{2n+1}-1\tag{1} \end{eqnarray*}$$ 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? $\endgroup$ – Jack D'Aurizio Jun 27 '17 at 19:40
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.