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)

  • $\begingroup$ wait, why do you have $-x_{2n-1}$. Aren't there absolute values? $\endgroup$ Jun 24, 2019 at 15:03
  • $\begingroup$ 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. $\endgroup$ Jun 24, 2019 at 15:05
  • $\begingroup$ @mathworker21 The minus sign is due to the absolute value. $\endgroup$ Jun 24, 2019 at 15:06
  • $\begingroup$ I've tried to prove monotonicity of $\{x_{2n}\}$, but I'm really not good at inequality. $\endgroup$ Jun 24, 2019 at 15:26

5 Answers 5


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.

About the monotonie.

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

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 $x\geq 0$:

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

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


$(\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 $\,a>-x$

  • $\begingroup$ Thanks for the explicit calculation! But this is not what I'm after (see the update part). $\endgroup$ Jun 25, 2019 at 9:10
  • $\begingroup$ @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. ;) $\endgroup$
    – user90369
    Jun 25, 2019 at 9:12
  • $\begingroup$ I think proving the monotonicity is harder than finding the limit. Why can't we try it? :) $\endgroup$ Jun 25, 2019 at 9:13
  • $\begingroup$ 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.) $\endgroup$
    – user90369
    Jun 25, 2019 at 9:20
  • $\begingroup$ 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. $\endgroup$ 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 $$ \lim_{n\to\infty}\frac1{\sqrt{n}}\left|\,\sum_{k=1}^n(-1)^k\sqrt{k}\,\right|=\frac12\tag7 $$

  • $\begingroup$ A nice demonstration of telescoping series and squeeze theorem! (+1) I also like the [a,b] notation. $\endgroup$ 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.
  • $\begingroup$ Thanks. Can you show me how to prove $\{x_{2n}\}$ is monotonically increasing without using integrals or something alike? (see my update) $\endgroup$ Jun 24, 2019 at 16:12
  • $\begingroup$ In point 2, $a_{2n}$ is written as sum of non-negative terms. $\endgroup$ Jun 24, 2019 at 16:13
  • $\begingroup$ You have the $(2n)^{-1/2} $ in front of the sum. But $(2n)^{-1/2}$ is decreasing. $\endgroup$ Jun 24, 2019 at 16:15
  • $\begingroup$ Oh, you are right. It is not that simple. $\endgroup$ 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.

  • $\begingroup$ 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) $\endgroup$
    – user90369
    Jun 25, 2019 at 20:48
  • $\begingroup$ Thanks for all the effort you've put in to answer the question. I haven't seen this technique before. $\endgroup$ Jun 26, 2019 at 13:01

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