# Prove $\lim\limits_{n\to \infty}\frac{1}{\sqrt n}\left|\sum\limits_{k=1}^n (-1)^k\sqrt k\right|= \frac{1}{2}$

I'm trying to show that $$\lim_{n\to \infty} x_n=\lim_{n\to \infty}\frac{1}{\sqrt n}\left|\sum_{k=1}^n (-1)^k\sqrt k\right|= \frac{1}{2}.$$ Assuming $$\lim\limits_{n\to\infty} x_n=x$$ exists, we have

$$x_{2n}=\frac{\sqrt{2n-1}(-x_{2n-1})+\sqrt {2n}}{\sqrt {2n}}$$

Letting $$n\to \infty$$,

$$\quad \quad x=-x+1$$ $$x=\frac{1}{2}$$

But I'm stuck on proving the existence of $$\lim x_n$$. Any idea?

Update: I just solved the problem using sandwich theorem + integral test. Still, I would like to see a continuation of my initial idea, i.e. proving

$$\{x_{2n}\}$$ is monotonically increasing (similarly, $$\{x_{2n+1}\}$$ is monotonically decreasing)

• wait, why do you have $-x_{2n-1}$. Aren't there absolute values? Jun 24, 2019 at 15:03
• anyways, you can show the limit exists by showing odd terms $x_{2n+1}$ are decreasing and the even ones $x_{2n}$ are increasing, which you can maybe show via the functional equation you have. Jun 24, 2019 at 15:05
• @mathworker21 The minus sign is due to the absolute value. Jun 24, 2019 at 15:06
• I've tried to prove monotonicity of $\{x_{2n}\}$, but I'm really not good at inequality. Jun 24, 2019 at 15:26

You can also apply Stolz–Cesàro, for $$z_{2n}=\frac{\sum\limits_{i=1}^{n}\left(\sqrt{2i}-\sqrt{2i-1}\right)}{\sqrt{2n}}=\frac{a_n}{b_n}$$ where $$b_n=\sqrt{2n}$$ is strictly monoton and divergent. Then $$\frac{a_{n+1}-a_n}{b_{n+1}-b_n}= \frac{\sqrt{2(n+1)}-\sqrt{2n+1}}{\sqrt{2(n+1)}-\sqrt{2n}}=\\ \frac{1}{2}\cdot \frac{\sqrt{2(n+1)}+\sqrt{2n}}{\sqrt{2(n+1)}+\sqrt{2n+1}}\to \frac{1}{2}, n\to\infty$$ and finally $$z_{2n+1}=\left|z_{2n}\cdot\sqrt{\frac{2n}{2n+1}}-1\right|\to\left|\frac{1}{2}-1\right|=\frac{1}{2}, n\to\infty$$

$$\displaystyle \lim_{n\to\infty}\frac{1}{\sqrt{n}}\left|\sum\limits_{k=1}^n (-1)^k\sqrt{k}\right| = \lim_{n\to\infty}\frac{1}{\sqrt{2n}}\sum\limits_{k=1}^n \frac{1}{\sqrt{2k-1}+\sqrt{2k}}$$

$$\displaystyle \frac{1}{2\sqrt{2n}}\sum\limits_{k=1}^n\frac{1}{ \sqrt{2k} } < \frac{1}{\sqrt{2n}}\sum\limits_{k=1}^n\frac{1}{\sqrt{2k-1} +\sqrt{2k} } \displaystyle < \frac{1}{2\sqrt{2n}}\sum\limits_{k=1}^n\frac{1}{\sqrt{2k-1} } < \frac{1}{2\sqrt{2n}}\left(1+\sum\limits_{k=1}^{n-1} \frac{1}{ \sqrt{2k} }\right)$$

With Riemann we get:

$$\displaystyle \frac{1}{\sqrt{2n}}\sum\limits_{k=1}^n\frac{1}{ \sqrt{2k} } = \frac{1}{2n}\sum\limits_{k=1}^n\frac{1}{\sqrt{k/n}} \to \frac{1}{2}\int\limits_0^1\frac{dx}{\sqrt{x}} = \sqrt{x}|_0^1 =1$$

And therefore the confirmation of the claim.

$$\displaystyle s_n := \sum\limits_{k=1}^n \frac{1}{\sqrt{2k-1}+\sqrt{2k}}$$

Assumption about the monotonie: $$\enspace \displaystyle \frac{s_n}{\sqrt{2n}} \enspace$$ is strictly increasing.

$$\displaystyle \frac{s_n}{\sqrt{2n}} < \frac{s_{n+1}}{\sqrt{2{n+2}}} \,$$ leads to

$$\displaystyle \left(\frac{1}{\sqrt{2n}} - \frac{1}{\sqrt{2n+2}}\right)s_n < \frac{1}{\sqrt{2n+2}}\frac{1}{ \sqrt{2n+1} + \sqrt{2n+2} }\,$$ and therefore to

$$\displaystyle s_n < a_n:=\frac{\sqrt{2n}}{2}\frac{ \sqrt{2n+2}+\sqrt{2n} }{\sqrt{2n+2} +\sqrt{2n+1} }$$

For $$n=1$$ it’s o.k. . Assume it’s correct for $$n$$ . $$(*)$$

Then we have to show that it's also correct for $$n \to n+1$$ .

$$\displaystyle s_{n+1} < \frac{\sqrt{2n}}{2}\frac{ \sqrt{2n+2}+\sqrt{2n} }{\sqrt{2n+2} +\sqrt{2n+1} } + \frac{1}{\sqrt{2n+1} +\sqrt{2n+2} } < a_{n+1}$$

The first inequation follows from the assumption $$(*)$$ and the second inequation can be proved by some transformations, $$2n$$ replaced by $$x$$ and $$x\geq 0$$:

$$(\sqrt{x}(\sqrt{x+2} +\sqrt{x}) + 2)(\sqrt{x+4}+\sqrt{x+3}) <$$

$$<\sqrt{x+2}(\sqrt{x+4}+ \sqrt{x+2})( \sqrt{x+2}+\sqrt{x+1})$$

Transformations:

$$(\sqrt{x}(\sqrt{x+2} +\sqrt{x}) + 2)(\sqrt{x+4}+\sqrt{x+2}) + (\sqrt{x}(\sqrt{x+2} +\sqrt{x}) + 2)(\sqrt{x+3}-\sqrt{x+2}) <$$ $$<\sqrt{x+2}(\sqrt{x+4}+ \sqrt{x+2})( \sqrt{x+2}+\sqrt{x+1})$$

$$(\sqrt{x}(\sqrt{x+2} +\sqrt{x}) + 2)(\sqrt{x+3}-\sqrt{x+2}) < (\sqrt{x+2}(\sqrt{x+2}+\sqrt{x+1})-(\sqrt{x}(\sqrt{x+2}+\sqrt{x})+2))(\sqrt{x+4}+\sqrt{x+2})$$

$$(\sqrt{x}\sqrt{x+2} +x+2)(\sqrt{x+3}-\sqrt{x+2}) < (\sqrt{x+2}\sqrt{x+4} +x+2)(\sqrt{x+1}-\sqrt{x})$$

This is true because of:

$$\sqrt{x}\sqrt{x+2} +x+2< \sqrt{x+2}\sqrt{x+4} +x+2$$

$$\sqrt{x+3}-\sqrt{x+2} < \sqrt{x+1}-\sqrt{x}$$

Note: $$\enspace\sqrt{x+1+a}-\sqrt{x+a}~$$ is decreasing by growing $$\,a>-x$$

• Thanks for the explicit calculation! But this is not what I'm after (see the update part). Jun 25, 2019 at 9:10
• @YuiToCheng : Yes, o.k., but it's the answer for the headline. And it's not so clear for me why you like to calculate more complicate then necessary. ;) Jun 25, 2019 at 9:12
• I think proving the monotonicity is harder than finding the limit. Why can't we try it? :) Jun 25, 2019 at 9:13
• That's what I mean: Why calculating harder then necessary ? Otherwise the headline is not correct, it would be better to write "problem with monotonie of ..." . And: Due to the limitation of the term (see inequalities) it's at least clear that the curve of the term is limited with increasing n. (If I find time I will think more about that.) Jun 25, 2019 at 9:20
• My initial question is to prove the limit in the headline. It is rude to change it IMO. But asking a follow-up question should be alright. Jun 25, 2019 at 9:23

Since $$\frac1{\sqrt{2k+1}+\sqrt{2k-1}}\le\frac1{\sqrt{2k}+\sqrt{2k-1}}\le\frac1{\sqrt{2k}+\sqrt{2k-2}}\tag1$$ we have $$\tfrac12\left(\sqrt{2k+1}-\sqrt{2k-1}\right)\le\sqrt{2k}-\sqrt{2k-1}\le\tfrac12\left(\sqrt{2k}-\sqrt{2k-2}\right)\tag2$$

Summing over an even number of terms gives \begin{align} \hspace{-1cm}\frac1{\sqrt{2n}}\sum_{k=1}^{2n}(-1)^k\sqrt{k} &=\frac1{\sqrt{2n}}\sum_{k=1}^n\left(\sqrt{2k}-\sqrt{2k-1}\right)\\ &=\frac1{\sqrt{2n}}\sum_{k=1}^n\left[\tfrac12\left(\sqrt{2k+1}-\sqrt{2k-1}\right),\tfrac12\left(\sqrt{2k}-\sqrt{2k-2}\right)\right]_\#\\ &=\frac1{\sqrt{2n}}\left[\,\tfrac12\left(\sqrt{2n+1}-1\right),\tfrac12\sqrt{2n}\,\right]_\#\tag3 \end{align} where $$[a,b]_\#$$ represents a number in $$[a,b]$$.

Therefore, by the Squeeze Theorem, $$\lim_{n\to\infty}\frac1{\sqrt{2n}}\sum_{k=1}^{2n}(-1)^k\sqrt{k}=\frac12\tag4$$

Summing over an odd number of terms gives \begin{align} \frac1{\sqrt{2n+1}}\sum_{k=1}^{2n+1}(-1)^k\sqrt{k} &=\frac1{\sqrt{2n+1}}\left(\sum_{k=1}^n\left(\sqrt{2k}-\sqrt{2k-1}\right)-\sqrt{2n+1}\right)\\ &=\frac1{\sqrt{2n+1}}\left[\,\tfrac12\left(\sqrt{2n+1}-1\right),\tfrac12\sqrt{2n}\,\right]_\#-1\tag5 \end{align} Therefore, by the Squeeze Theorem, $$\lim_{n\to\infty}\frac1{\sqrt{2n+1}}\sum_{k=1}^{2n+1}(-1)^k\sqrt{k}=-\frac12\tag6$$

Combining $$(4)$$ and $$(6)$$ yields $$\lim_{n\to\infty}\frac1{\sqrt{n}}\left|\,\sum_{k=1}^n(-1)^k\sqrt{k}\,\right|=\frac12\tag7$$

• A nice demonstration of telescoping series and squeeze theorem! (+1) I also like the [a,b] notation. Jun 27, 2019 at 3:52

Here are some steps.

1. Let $$a_n:=n^{-1/2}\sum_{k=1}^n(-1)^k\sqrt k$$. Study separately the behavior of $$(a_{2n})$$ and $$(a_{2n+1})$$.
2. For $$a_{2n}$$: notice that this term is equal to $$(2n)^{-1/2}\sum_{j=1}^n\left(\sqrt{2j}-\sqrt{2j-1}\right)=(2n)^{-1/2}\sum_{j=1}^n\frac{1}{\sqrt{2j}+\sqrt{2j-1}}.$$
3. Add and substract $$(2n)^{-1/2}\sum_{j=1}^n\frac{1}{2\sqrt{2j} }$$ and show that $$(2n)^{-1/2}\sum_{j=1}^n\frac{1}{\sqrt{2j}+\sqrt{2j-1}} -(2n)^{-1/2}\sum_{j=1}^n\frac{1}{2\sqrt{2j} }$$ goes to zero.
4. It remains to study the limit of $$4^{-1}n^{-1/2}\sum_{k=1}^nk^{-1/2}$$ for example comparing with an integral.
• Thanks. Can you show me how to prove $\{x_{2n}\}$ is monotonically increasing without using integrals or something alike? (see my update) Jun 24, 2019 at 16:12
• In point 2, $a_{2n}$ is written as sum of non-negative terms. Jun 24, 2019 at 16:13
• You have the $(2n)^{-1/2}$ in front of the sum. But $(2n)^{-1/2}$ is decreasing. Jun 24, 2019 at 16:15
• Oh, you are right. It is not that simple. Jun 24, 2019 at 16:16

Sums like this can be solved in general by the method of analytic regularization i.e.: From $$z^{-\epsilon} = \frac{1}{\Gamma(\epsilon)} \int_0^\infty \frac{{\rm d}t}{t} \, t^{\epsilon} \, {\rm e}^{-zt}$$ for $$z=\frac{k}{n}$$ one obtains $$\sum_{k=1}^{n} (-1)^k \left(\frac{k}{n}\right)^{-\epsilon} = \frac{1}{\Gamma\left(\epsilon\right)} \int_0^\infty \frac{{\rm d}t}{t} \, t^{\epsilon} \sum_{k=1}^{n} (-1)^k \, {\rm e}^{-{\frac{k}{n}}\,t} = \frac{1}{\Gamma\left(\epsilon\right)} \int_0^\infty \frac{{\rm d}t}{t} \, t^{\epsilon} \, \frac{(-1)^n \, {\rm e}^{-t}-1}{{\rm e}^{{t}/{n}}+1}$$ This line is fully valid for all $$\epsilon>0$$, in which case however, when taking the limit $$n\rightarrow \infty$$, the LHS manifestly diverges. Nevertheless, when $$\epsilon<0$$ the RHS can be understood as a to be regularized integral, meaning if it acquires a value it must correspond to a value of the LHS for $$\epsilon<0$$. The last equality also shows that one has to differ between even and odd $$n$$. When making a choice for either, the limit $$n \rightarrow \infty$$ exists and the value can be recovered by analytic regularization. Hence: $$\stackrel{n \rightarrow \infty}{=} \frac{1}{2\,\Gamma\left(\epsilon\right)} \int_0^\infty \frac{{\rm d}t}{t} \, t^{\epsilon} \left\{(-1)^n \, {\rm e}^{-t}-1 \right\} = \frac{(-1)^n\Gamma\left(\epsilon\right) - 0}{2\,\Gamma\left(\epsilon\right)} = \frac{(-1)^n}{2}$$ where the $$(-1)^n$$ stands symbolically for the choice of even or odd $$n$$. The second integral is zero in terms of analytic continuation, since it is a pure power (Veltman rule).

monotonicity: For even $$n$$ the above integral is convergent even for $$-1<\epsilon<0$$ in which case $$\Gamma(\epsilon)<0$$ and the integral then represents the continuous version of the discret LHS for even $$n$$. The derivative of the integrand with respect to $$n$$ (once the even branch is chosen: $$n\rightarrow 2n$$) $$\frac{t^\epsilon \left(1-{\rm e}^{-t}\right) {\rm e}^{\frac{t}{2n}}}{-2n^2\,\Gamma(\epsilon) \left( {\rm e}^{\frac{t}{2n}} + 1 \right)^2}$$ is manifestly positive.

• Yes, analytical regularization is a nice way to calculate, but I think the OP would like to have it easier. But nevertheless: A good method. (+1) Jun 25, 2019 at 20:48
• Thanks for all the effort you've put in to answer the question. I haven't seen this technique before. Jun 26, 2019 at 13:01