# Solve limit using definite integration

I have to solve this limit using definite integral

$${\lim_{n\to \infty} \frac{1}{n} \cdot \sum_{i=1}^n \frac{1}{1+(\frac{i}{n})^i}}$$

Well, my development was:

I tried to give the Riemann integral definition form, using a regular partition such that:

$$\int_a^bf(x)dx=\lim_{n\to\infty}\frac{b-a}{n}\cdot\sum_{i=1}f\left(a+i\frac{b-a}{n}\right)$$

So, i set $$b=1, a=0$$ for simplicity, then we have $$f(\frac{i}{n})$$ must be equal to $${\frac{1}{1+(\frac{i}{n})^i}}$$.

Let $${x=\frac{i}{n}}$$, then $${f(x)=\frac{1}{1+x^{nx}}}$$

The problem is that $$n$$ is a dummy variable that does not make sense outside the limit, that is, it does not make sense for the function $$f$$

So, I need to do some kind of algebraic transformation or variable change for $$n$$, in such a way that the function $$f$$ remains only in terms of $$x$$ and thus I can use it in the definite integral. However, I have not been able to find such a magical algebraic transformation or variable change for $$n$$.

• I recommend using squeeze theorem. Can you bound the summand from above and below while getting rid of the exponent? Hint: $\frac{i}{n}\leq 1$ Dec 19, 2020 at 18:24
• Do you have to do it using Riemann sums? It seems easier to bound the sum then note that it is basically a Cesaro sum. Dec 19, 2020 at 19:52
• Yes, only Riemann sums Dec 19, 2020 at 20:02
• @NinadMunshi Can you provide more hint? Dec 19, 2020 at 20:05
• @EduardoSebastian I don't think you can do this just directly using Riemann integration. You need $\frac{1}{1+(i/n)^i}$ to be a function of solely $i/n$, which, as you noted, it is not. Dec 31, 2020 at 18:45

Denote the expression under the limit by $$a_n$$. Clearly $$a_n<1$$, so that $$\limsup\limits_{n\to\infty}a_n\leqslant 1$$.

Further, for a fixed $$n$$, and positive real values of $$x$$, the map $$x\mapsto(x/n)^x$$ has a global minimum at $$x=n/e$$. Thus, for $$1\leqslant i\leqslant m$$ we have $$(i/n)^i\leqslant\max\{1/n,(m/n)^m\}$$.

This gives an idea for a lower bound. Let $$0<\varepsilon<1$$. Then $$1\leqslant i\leqslant(1-\varepsilon)n\implies(i/n)^i\leqslant\max\{1/n,(1-\varepsilon)^{n(1-\varepsilon)}\}\color{gray}{\underset{n\to\infty}{\longrightarrow}0},$$ hence there exists $$N_\varepsilon$$ such that for $$n>N_\varepsilon$$ and $$1\leqslant i\leqslant(1-\varepsilon)n$$ we have $$(i/n)^i<\varepsilon$$, and $$a_n>\frac1n\times\lfloor(1-\varepsilon)n\rfloor\times\frac1{1+\varepsilon}\implies\liminf_{n\to\infty}a_n\geqslant\frac{1-\varepsilon}{1+\varepsilon}.$$

Since $$\varepsilon$$ is arbitrary, we have $$\liminf\limits_{n\to\infty}a_n\geqslant 1$$, giving finally $$\lim\limits_{n\to\infty}a_n=1$$.

A variant with Riemann sums (far-fetched): take a positive integer $$m$$, then for $$n\geqslant m$$ $$a_n:=\frac1n\sum_{i=1}^n\frac{1}{1+(i/n)^i}\geqslant\frac1n\sum_{i=1}^nf_m(i/n),\quad f_m(x):=\frac{1}{1+x^{mx}},$$ so that $$\liminf\limits_{n\to\infty}a_n\geqslant\int_0^1 f_m(x)\,dx$$ and, since $$m$$ is arbitrary, we may take $$m\to\infty$$: $$\liminf_{n\to\infty}a_n\geqslant\lim_{m\to\infty}\int_0^1 f_m(x)\,dx=1.$$

• Thanks for your help, but i need to use Riemann integration Dec 22, 2020 at 16:54
• @EduardoSebastian: added a "Riemannian" approach. (It's not uncommon to see a "Riemann-sum-free" solution to a "Riemann-sum-like" problem: here is an example where both approaches are possible, and here is an example where I just can't see Riemann sums work.) Dec 23, 2020 at 8:37

As it is really shown in OP, for $$\;x=\dfrac in\in(0,1]\;$$ $$f(x) =\lim\limits_{n\to\infty}\;\dfrac 1{1+x^{nx}}=\genfrac{\{}.0{}{1,\text{ if }x\in(0,1)}{\;^1/_2,\text{ if } \;x=1,\;}$$ i.e. the function $$\;f(x)\;$$ has a removable discontinuity at the point $$\;x=1\;$$ and is equaled to $$1$$ at the other points of the domain.

Therefore, $$\color{brown}{\mathbf{\lim\limits_{n\to\infty}\,\dfrac1n\,\dfrac1{1+\left(\large\frac in\right)^i} = \int\limits_0^1\, 1\,\text dx =1.}}$$

• $x$ is related to $n$, for example, $x = \frac{n-1}{n}$ (when $i=n-1$), $\lim\limits_{n\to\infty}\;\dfrac 1{1+x^{nx}} = \frac{1}{1+ \mathrm{e}^{-1}}$, right? Jan 4, 2021 at 12:11
• @RiverLi $\;x=\frac{n-1}n\;$ means $\;x=1.$ Jan 4, 2021 at 13:22
• Can you explain it in detail? Jan 4, 2021 at 13:38
• @RiverLi $\;\lim\limits_{n\to \infty}\frac{n-1}n=1.\;$ :-) Jan 4, 2021 at 13:46
• But $x$ depends on $n$, $\lim_{n\to \infty} x^{nx} = \lim_{n\to \infty} (\frac{n-1}{n})^{n-1} = \mathrm{e}^{-1}$. Jan 4, 2021 at 14:04

By proceeding through your method, we have obtained f(x) as $$\frac{1}{1+x^{nx}}$$, as mentioned in the question.

So, when n approaches infinity,

$$f(x)=0$$ if $$x$$ $$ϵ$$ $$(0,1),$$

and at $$x=1$$, $$lim$$ $$n->∞, f(x)=2$$

So, $$f(x)=1$$ at all but finite points and hence,

Given Sum= $$\int_{0}^{1}1dx$$

$$=1$$

(We can further write the integral as $$\int_{0}^{1}1dx$$ + $$\int_{1}^{1}2dx$$ = $$1+0$$ but that is understood.)