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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.

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6 Answers 6

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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 $$\lim_{n\to\infty} n \sum_{k=n}^\infty\frac{1}{2k(2k+1)} = \frac14$$

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  • $\begingroup$ Lovely! I made a similar squeeze theorem argument, but with the integral bounds of the tail series. However, this proof is much more elegant. Thanks! $\endgroup$
    – srnoren
    Jun 15, 2015 at 7:29
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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,

$$\lim_{n \to \infty} n \sum_{k=n}^{\infty} \frac1{2 k (2 k+1)} = \frac14 $$

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Hint: $$\frac1{2k(2k+1)} = \frac1{2k} - \frac1{2k+1}.$$

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    $\begingroup$ But you don't get all the terms canceling. $\endgroup$ Jun 15, 2015 at 5:23
  • $\begingroup$ That was one of my first ideas, but it doesn't simplify my expression for the series tails. It does, however, change the tails to those of an alternating harmonic series: $$\sum^\infty_{k=2n} \frac{(-1)^k}{k}.$$ Is the order of this sequence known? $\endgroup$
    – srnoren
    Jun 15, 2015 at 5:39
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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 $$\lim_{n\rightarrow\infty}n\sum\limits_{k=n}^\infty\frac{1}{2k(2k+1)}=\lim_{n\rightarrow\infty}\frac{n+1}{4n+2}=\frac{1}{4}.$$

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    $\begingroup$ The equality of two limits does not imply that the sequences have the same order. For example, $$\lim_{n\rightarrow \infty} \frac1{n^{1000}}=0=\lim_{n\rightarrow \infty} \frac1{n}.$$ If I multiplied $n$ inside both limits, I'd get different results. $\endgroup$
    – srnoren
    Jun 15, 2015 at 5:53
  • $\begingroup$ ah. excellent point. Informally speaking, the summation tends to add 1 to the order, and so i would guess the two sequences do indeed have the same order; however, proving that is probably just unnecessarily complicating things :/ $\endgroup$
    – Brent
    Jun 15, 2015 at 6:10
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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, $$nS = \frac{1}{4}$$.

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  • $\begingroup$ $\lim_n S_n = 0$. But you cannot get $S = \frac{1}{2n} - S$ as you need to take $\frac{1}{2n}\to 0$ too. $\endgroup$
    – user99914
    Jun 15, 2015 at 5:52
  • $\begingroup$ Take the sum up to $N$ instead of $\infty$, and see what happens as $N$ gets large. I think your way will work. $\endgroup$ Jun 15, 2015 at 5:52
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    $\begingroup$ To repair and complete this approach, consider $$S_n=\sum_{k=n}^{\infty}\frac{1}{2k(2k+1)}\qquad T_n=\sum_{k=n}^{\infty}\frac{1}{(2k+1)(2k+2)}$$ then $$\frac{n}{n+1}S_n\leqslant T_n\leqslant S_n$$ hence $$\lim\limits_{n\to\infty}nS_n=\lim\limits_{n\to\infty}nT_n.$$ Since (as shown in the answer) $S_n+T_n=1/(2n)$, this proves that $$\lim\limits_{n\to\infty}nS_n=\lim\limits_{n\to\infty}nT_n=\frac14.$$ $\endgroup$
    – Did
    Jun 15, 2015 at 6:12
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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.

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