Dear Sirs and Dear Madams,
I need to calculate the limit of $$S_{n} = \sum\limits_{k =1}^{n} [ \dfrac{1}{4k-3} + \dfrac{1}{4k - 1} - \dfrac{1}{2k} ] .$$
I show that \begin{eqnarray*} \lim\limits_{n \to +\infty } S_{n} &=& \lim\limits_{n\to +\infty} \sum\limits_{k = 1}^{n} \left[ \dfrac{1}{4k-3} - \dfrac{1}{4k-2} + \dfrac{1}{4k-1} - \dfrac{1}{4k} + \dfrac{1}{2} \left[ \dfrac{1}{2k-1} -\dfrac{1}{2k} \right] \right] \\ &=& \dfrac{3}{2} \lim\limits_{n \to + \infty} \sum\limits_{k = 1}^{n} \left[ \dfrac{1}{2k-1} -\dfrac{1}{2k} \right] \\ &=& \dfrac{3}{2} \ln 2. \end{eqnarray*}
However, I tried to do that by using "Euler-Maclaurin summation formula" as follows:
- We have $ S_{n} = \sum\limits_{k =1}^{n} (\dfrac{1}{4k-3} + \dfrac{1}{4k-1} - \dfrac{1}{2k})$.
Put \begin{eqnarray*} f(x) = \dfrac{1}{4x-3} + \dfrac{1}{4x-1} - \dfrac{1}{2x}. \end{eqnarray*} Then \begin{eqnarray} f^{(k)} (x) = \dfrac{(-4)^k k!}{(4x-3)^{k+1}} + \dfrac{(-4)^k k!}{(4x-1)^{k+1}} - \dfrac{(-1)^k k!}{2(x)^{k+1}} \end{eqnarray} 2. Using Euler's formula, we have \begin{eqnarray} S_{n} &=& \int\limits_1^n f(x) dx + \sum\limits_{k=1}^m (-1)^k \dfrac{B_k}{k!} \left\lbrace f^{(k-1)}(N) - f^{k-1}(1) \right\rbrace + \dfrac{(-1)^{m-1}}{m!} \int\limits_1^{N} \overline{B_m}(x) f^{(m)} (x) dx. \end{eqnarray} Choose $m = 1$ and note that $\overline{B_1}(x) = x - [x] - \dfrac{1}{2}$, we have \begin{eqnarray} S_{n} &=& \int\limits_1^n f(x) dx - \dfrac{B_1}{1!} \left\lbrace f(n) - f(1) \right\rbrace + \dfrac{1}{1!} \int\limits_1^{n} \overline{B_1}(x) f^{'} (x) dx.\\ &=& \int\limits_1^n f(x) dx - \dfrac{B_1}{1!} \left\lbrace f(n) - f(1) \right\rbrace + \dfrac{1}{1!} \int\limits_1^{n} (x - [x] - \dfrac{1}{2}) f^{'} (x) dx.\\ &=& \int\limits_1^n f(x) dx + \dfrac{1}{2} \left\lbrace f(n) - f(1) \right\rbrace + \int\limits_1^{n} x f^{'} (x) dx - \dfrac{1}{2} \int\limits_1^{n} f^{'} (x) dx - \int\limits_1^{n} [x] f^{'} (x) dx.\\ &=& \int\limits_1^n f(x) dx + \int\limits_1^{n} x f^{'} (x) dx - \int\limits_1^{n} [x] f^{'} (x) dx.\\ &=& \int\limits_1^n f(x) dx + nf(n) - f(1) - \int\limits_1^{n} f (x) dx - \int\limits_1^{n} [x] f^{'} (x) dx.\\ &=& nf(n) - f(1) - \int\limits_1^{n} [x] f^{'} (x) dx.\\ &=& \dfrac{n}{4n-3} + \dfrac{n}{4n-1} - \dfrac{1}{2} - 1 - \dfrac{1}{3} + \dfrac{1}{2} - \sum\limits_{k=1}^{n-1} k \int\limits_k^{k+1} f^{'} (x) dx.\\ &=& \dfrac{n}{4n-3} + \dfrac{n}{4n-1} - \dfrac{4}{3} - \sum\limits_{k=1}^{n-1} k (f(k+1) - f(k))\\ &=& \dfrac{n}{4n-3} + \dfrac{n}{4n-1} - \dfrac{4}{3} - \sum\limits_{k=1}^{n-1} \{ \dfrac{k}{4k+1} + \dfrac{k}{4k+3} - \dfrac{k}{2k+2} - \dfrac{k}{4k-3} - \dfrac{k}{4k-1} + \dfrac{1}{2} \}\\ &=& \dfrac{n}{4n-3} + \dfrac{n}{4n-1} - \dfrac{4}{3} - \sum\limits_{k=1}^{n-1} \{ - \dfrac{1}{4(4k+1)} - \dfrac{1}{4(4k+3)} + \dfrac{1}{2k+2} - \dfrac{1}{4(4k-3)} - \dfrac{1}{4(4k-1)} \}\\ \end{eqnarray} 3. Takes $n \to +\infty $, we obtain that \begin{eqnarray} \lim\limits_{n\to +\infty} S_n &=& \dfrac{1}{4} + \dfrac{1}{4} - \dfrac{4}{3} + \sum\limits_{k=1}^{+\infty} \{ \dfrac{1}{4(4k+1)} + \dfrac{1}{4(4k+3)} - \dfrac{1}{2k+2} + \dfrac{1}{4(4k-3)} + \dfrac{1}{4(4k-1)} \}\\ &=& -\dfrac{5}{6} + \dfrac{1}{4} \sum\limits_{k=1}^{+\infty} \{ \dfrac{1}{4k+1} + \dfrac{1}{4k+3} - \dfrac{2}{k+1} + \dfrac{1}{4k-3} + \dfrac{1}{4k-1} \}. \end{eqnarray} My question is that "How do I do to get to the last result (i.e. $\dfrac{3}{2} \ln 2$)?"
Thank you very much for your important helps?