# Show that $\lim_{n\rightarrow \infty }\frac{1}{n}\sum_{k=1}^{n}(-1)^kf\left ( \frac{k}{n} \right )=0$

Let $$f : [0, 1] → \mathbb{R}$$ be a continuous function. Prove that $$\lim_{n\rightarrow \infty }\frac{1}{n}\sum_{k=1}^{n}(-1)^kf\left ( \frac{k}{n} \right )=0$$

First, I observed that any pair of consecutive terms' input values have the same distance $$\frac{1}{n}$$, and this converges to $$0$$ as $$n$$ goes to infinity.

Proof:

Consider $$f(\frac{a}{n})$$ and $$f(\frac{a+1}{n})$$ for any $$a\in \{1,2,3,...,n\}$$.

Note that for any $$a$$, $$\lim_{n\rightarrow \infty }(\frac{a+1}{n}-\frac{a}{n})=\lim_{n\rightarrow \infty }(\frac{1}{n})=0$$.

So, for some $$\delta_{1}>0,\; \; \; \exists N$$ s.t $$\left | \frac{1}{n} \right |< \delta _{1}\; \; \; \forall n>N$$

And since $$f$$ is continuous at $$\frac{a}{n}$$, $$\forall \varepsilon > 0, \; \; \; \exists 0<\delta<\delta _{1}$$ s.t $$\forall x\in [0,1] \; \; \; \text {that satisfies}\; \; \;\left | x-\frac{a}{n} \right |< \delta < \delta _{1}, \text {we have}\; \; \;\left | f(x)-f(\frac{a}{n}) \right |< \varepsilon$$

Thus $$\forall n>N$$, we get $$\left |\sum_{k=1}^{n}(-1)^kf(\frac{k}{n}) \right |=\left | -f(\frac{1}{n})+f(\frac{2}{n})-\cdot \cdot \cdot +(-1)^{n-1}f(\frac{n-1}{n})+(-1)^nf(\frac{n}{n}) \right |< \frac{n}{2}\varepsilon$$

This is because we have at least $$\frac{2}{n}$$ many pairs of consecutive terms, and we can bound the series by Triangle Inequality.

Then this implies, $$\frac{1}{n}\left |\sum_{k=1}^{n}(-1)^kf(\frac{k}{n}) \right |= \left |\frac{1}{n} \right |\left |\sum_{k=1}^{n}(-1)^kf(\frac{k}{n}) \right |<\frac{\varepsilon }{2}<\varepsilon$$.

Hence, for $$n>N$$ and for any $$\varepsilon$$, $$\left |\frac{1}{n} \sum_{k=1}^{n}(-1)^kf(\frac{k}{n}) \right |<\varepsilon$$, so $$\lim_{n\rightarrow \infty }\frac{1}{n}\sum_{k=1}^{n}(-1)^kf\left ( \frac{k}{n} \right )=0$$. $$\blacksquare$$

Am I missing something in the proof?

Also, I doubt if we really can find some $$0<\delta<\delta _{1}$$. In other words, if we are looking for some $$\delta$$ for a fixed $$\varepsilon$$, is $$\delta$$ always bounded?

• I think there is some mistake. Note that the $n$ is running, and you said that $f$ is continuous at $a/n$, for then the $\delta$ you choose afterward all depend on $n$, it seems that all are running, this is weird. May 7, 2021 at 12:47
• Express $\int _0^1 f(x) \, dx$ as limit of a Riemann sum using. Combining this Riemann sum with the sum given in question gives another sum which is also a Riemann sum for $f$ on $[0,1]$. Thus the desired limit is $0$. You can that we only need Riemann integrability of $f$ and not necessarily continuity. May 7, 2021 at 12:54
• math.stackexchange.com/q/2886467/42969 May 7, 2021 at 13:11
• @user284331 Then, can I just change the order of the proof? I mean, we can find some delta for any epsilon by using the continuity first, and then we can find N for such delta, then this mean, for n>N, any consecutive terms satisfies the $|x-a/n|$ so they are all bounded by epsilon. Do you think this works?
– john
May 7, 2021 at 13:34
• I think you really need to use uniform continuity to get rid of the particular choice of $a/n\in[0,1]$. May 7, 2021 at 13:35

$$1/n \sum_{k=1}^n (-1)^k f(k/n) + 1/n \sum_{k=1}^n f(k/n) = 2/n \sum_{k=1}^{n/2} f(2k/n) = 2/n \sum_{k=1}^{n/2} f(2k/n)$$ Now limit $$n \rightarrow \infty$$ => $$\lim_{n \rightarrow \infty }1/n \sum_k (-1)^k f(k/n) = \int_{0}^1 f(x) dx - \int_{0}^1 f(x) dx = 0$$ because $$\lim_{n \rightarrow \infty} 2/n \sum_{k=1}^{n/2} f(2k/n) = \lim_{n \rightarrow \infty} 1/n \sum_{k=1}^{n} f(k/n)=\int_{0}^1 f(x) dx$$

• Please fix your latex for the limit notation. May 7, 2021 at 13:00
• Thanks...fixed it. May 7, 2021 at 13:02
• +1 there. This is what I had suggested in comments and is the most straightforward approach to the problem. May 7, 2021 at 13:03

Since $$f$$ is uniformly continuous on $$[0,1]$$, for $$\epsilon>0$$, there is some $$\delta>0$$ such that $$|f(x)-f(y)|<\epsilon$$ for $$|x-y|<\delta$$.

We first look at the subsequence of even terms, that is, \begin{align*} \dfrac{1}{2n}\sum_{k=1}^{2n}(-1)^{k}f\left(\dfrac{k}{2n}\right) \end{align*} Choose an $$N$$ such that $$1/N<\delta$$. For $$n\geq N$$, we have \begin{align*} \left|\dfrac{1}{2n}\sum_{k=1}^{2n}(-1)^{k}f\left(\dfrac{k}{2n}\right)\right|\leq\dfrac{1}{2n}\sum_{k=1}^{n}\left|f\left(\dfrac{k+1}{2n}\right)-f\left(\dfrac{k}{2n}\right)\right|\leq\dfrac{1}{2n}\sum_{k=1}^{n}\epsilon<\dfrac{\epsilon}{2}. \end{align*} For the subsequence of odd terms, one sees that

\begin{align*} \dfrac{1}{2n+1}\sum_{k=1}^{2n+1}(-1)^{k}f\left(\dfrac{k}{2n}\right)=-\dfrac{1}{2n+1}f\left(\dfrac{1}{2n+1}\right)+\dfrac{1}{2n+1}\sum_{k=2}^{2n+1}(-1)^{k}f\left(\dfrac{k}{2n+1}\right), \end{align*} and perform the similar trick for the second expression above since there are even terms for that.

• Can I get some hint for dealing with $-\dfrac{1}{2n+1}f\left(\dfrac{1}{2n+1}\right)$? I cannot get rid of it at the end..
– john
May 7, 2021 at 14:52
• That is easy since $f$ is continuous at $x=0$, so $f(1/(2n+1))\rightarrow 0$. May 7, 2021 at 14:54
• So, here we are observing two cases when n is even and when n is odd because we need to show the limit of the "$\frac{1}{n}$series" has to be bounded by epsilon for any n>N?
– john
May 7, 2021 at 14:57
• Not like that. It is merely because when you do the grouping inside the sum, you need even terms. May 7, 2021 at 14:59