$\lim_{n\rightarrow \infty} n \sum^\infty_{k=n} \frac1{2k(2k+1)}=\frac14$? This isn't a homework problem, just something that came up while I was studying measure theory. It is well known that the limit of the tails of any convergent series goes to 0. However, the problem that I have asks for the order in which such a limit vanishes. In particular, I wish to find $$\lim_{n\rightarrow \infty} n \sum^\infty_{k=n} \frac1{2k(2k+1)}.$$
Some playing around on Mathematica suggests that the limit should be $\frac14$. However, I'm having the worst time trying to show this with a straight-forward proof. As far as I can tell, there is no way to rewrite the tail series in terms of elementary functions of $n$ through the method of telescoping series or other similar series tricks. Am I missing something obvious, or is this limit really a bear to work through? I'd appreciate any help offered.
 A: Notice
$$\begin{array}{ccccc}
\frac12\left(\frac{1}{2k-1} - \frac{1}{2k+1}\right)&
& & &
\frac12\left(\frac{1}{2k} - \frac{1}{2k+2}\right)
\\
|| & & & & || \\
\frac{1}{(2k-1)(2k+1)} 
& \ge & 
\frac{1}{2k(2k+1)} 
& \ge &
\frac{1}{2k(2k+2)}\end{array}$$
The partial sums start at $k = n$ is squeezed between two telescoping series.
This leads to
$$\frac{1}{4n-2} \ge \sum_{k=n}^\infty\frac{1}{2k(2k+1)} \ge \frac{1}{4n}
$$
As a result,
$$ \left| n\sum_{k=n}^\infty\frac{1}{2k(2k+1)} - \frac14 \right| \le \frac{1}{8n-4}$$
Since $\displaystyle\;\lim_{n\to\infty} \frac{1}{8n-4} = 0$, we get 
$$\lim_{n\to\infty} n \sum_{k=n}^\infty\frac{1}{2k(2k+1)} = \frac14$$
A: This solution is based on the decomposition $$\frac1{2 k (2 k+1)} = \frac1{2 k} - \frac1{2 k+1} $$
and on the expansion up to order $\frac1n$ of the $n$th harmonic number $H_n$ as $$H_n=\sum_{k=1}^n\frac1k=\gamma+\log n+\frac1n+O\left(\frac1{n^2}\right).$$ 
Here we go:
$$\begin{align}\sum_{k=n}^{\infty} \frac1{2 k (2 k+1)} &= \frac1{2 n} - \frac1{2 n+1} +\frac1{2 n+2}-\frac1{2 n+3}+\cdots \\ &= \sum_{k=1}^{\infty} \frac{(-1)^{k}}{k} - \sum_{k=1}^{2 n-1} \frac{(-1)^{k}}{k}\\ &= \sum_{k=1}^{\infty} \frac{(-1)^{k}}{k} + \sum_{k=1}^{2 n-1} \frac{1}{k}-2 \sum_{k=1}^{ n-1} \frac{1}{2k}\\ &= -\log{2} + H_{2 n-1} - H_{n-1} \\ &= -\log{2} +\left (\gamma +  \log{(2 n-1)}+ \frac1{4 n-2}\right ) - \left (\gamma+\log{(n-1)} + \frac1{2 n-2} \right ) +O\left (\frac1{n^2}\right)\\ &= -\log{2} + \log{\left ( \frac{2 n-1}{n-1} \right )} + \frac1{4 n} - \frac1{2 n}+O\left (\frac1{n^2}\right)\\ &= -\log{2} + \log{2}+\log{\left ( 1+\frac1{2 (n-1)} \right )} -\frac1{4 n}+O\left (\frac1{n^2}\right)\\ &= \frac{1}{4 n}+O\left (\frac1{n^2}\right)\end{align}$$
Thus,
$$\lim_{n \to \infty} n \sum_{k=n}^{\infty} \frac1{2 k (2 k+1)} = \frac14 $$
A: Hint: $$\frac1{2k(2k+1)} = \frac1{2k} - \frac1{2k+1}.$$
A: If we consider that expression as a sequence, $a_n$, and if, as you say, the limit does indeed exist, then we know that $\lim(a_n)=\lim(a_{n+1})$. Hence,
$$\lim_{n\rightarrow\infty}n\sum\limits_{k=n}^\infty\frac{1}{2k(2k+1)}=\lim_{n\rightarrow\infty}(n+1)\sum\limits_{k=n+1}^\infty\frac{1}{2k(2k+1)}$$
$$=\lim_{n\rightarrow\infty}(n+1)\left\{\sum\limits_{k=n}^\infty\frac{1}{2k(2k+1)} -\frac{1}{2n(2n+1)}\right\} $$
$$ =\lim_{n\rightarrow\infty}n\sum\limits_{k=n}^\infty\frac{1}{2k(2k+1)}+\lim_{n\rightarrow\infty}\sum\limits_{k=n}^\infty\frac{1}{2k(2k+1)}-\lim_{n\rightarrow\infty}\frac{n+1}{2n(2n+1)}. $$
However, this implies that 
$$\lim_{n\rightarrow\infty}\sum\limits_{k=n}^\infty\frac{1}{2k(2k+1)}=\lim_{n\rightarrow\infty}\frac{n+1}{2n(2n+1)}$$
Multiplying by $n$, we get that 
$$\lim_{n\rightarrow\infty}n\sum\limits_{k=n}^\infty\frac{1}{2k(2k+1)}=\lim_{n\rightarrow\infty}\frac{n+1}{4n+2}=\frac{1}{4}.$$
A: Let $$S=\sum_{k=n}^{\infty}\frac{1}{2k(2k+1)}=\frac{1}{2n}-\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{2n+3}+...
$$
Note that
$$S=\frac{1}{2n}-\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{2n+3}+... $$
$$=\frac{1}{2n}-(\frac{1}{2n+1}-\frac{1}{2n+2})-(\frac{1}{2n+3}-\frac{1}{2n+4})-... 
$$
$$=\frac{1}{2n}-\frac{1}{(2n+1)(2n+2)}-\frac{1}{(2n+3)(2n+4)}-...$$
$$=\frac{1}{2n}-\sum_{k=n}^{\infty}\frac{1}{(2k+1)(2k+2)}$$
This part, I'm not 100% confident on, but I believe that 
$$\lim_{n \rightarrow \infty} S =  \lim_{n \rightarrow \infty}( \sum_{k=n}^{\infty}\frac{1}{(2k+1)(2k+2)})$$
In other words, as $n\rightarrow \infty$: $$S= \frac{1}{2n}-S$$  and
$$S = \frac{1}{4n}$$
Thus, $$nS = \frac{1}{4}$$.
A: The sum is less than
$$\sum_{k=n}^{\infty} \frac{1}{4k^2} < \int_{n-1}^\infty \frac{dx}{4x^2}= \frac{1}{4(n-1)}.$$
Similarly, the sum is greater than
$$\sum_{k=n}^{\infty} \frac{1}{(2k+1)^2}>\int_{n}^\infty \frac{dx}{(2x+1)^2}= \frac{1}{2(2n+1)}.$$
Multiplying by $n$ shows your expression is betweem $n/[2(2n+1)]$ and $n/[4(n-1)].$ Both of these $\to 1/4,$ and that is our limit.
