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:


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 $n$.

  • 5
    $\begingroup$ I recommend using squeeze theorem. Can you bound the summand from above and below while getting rid of the exponent? Hint: $\frac{i}{n}\leq 1$ $\endgroup$ Dec 19, 2020 at 18:24
  • $\begingroup$ 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. $\endgroup$ Dec 19, 2020 at 19:52
  • $\begingroup$ Yes, only Riemann sums $\endgroup$ Dec 19, 2020 at 20:02
  • $\begingroup$ @NinadMunshi Can you provide more hint? $\endgroup$ Dec 19, 2020 at 20:05
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    $\begingroup$ @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. $\endgroup$ Dec 31, 2020 at 18:45

3 Answers 3


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 $m\to\infty$: $$\liminf_{n\to\infty}a_n\geqslant\lim_{m\to\infty}\int_0^1 f_m(x)\,dx=1.$$

  • $\begingroup$ Thanks for your help, but i need to use Riemann integration $\endgroup$ Dec 22, 2020 at 16:54
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    $\begingroup$ @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.) $\endgroup$
    – metamorphy
    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, $$\color{brown}{\mathbf{\lim\limits_{n\to\infty}\,\dfrac1n\,\dfrac1{1+\left(\large\frac in\right)^i} = \int\limits_0^1\, 1\,\text dx =1.}}$$

  • $\begingroup$ $x$ 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? $\endgroup$
    – River Li
    Jan 4, 2021 at 12:11
  • $\begingroup$ @RiverLi $\;x=\frac{n-1}n\;$ means $\;x=1.$ $\endgroup$ Jan 4, 2021 at 13:22
  • $\begingroup$ Can you explain it in detail? $\endgroup$
    – River Li
    Jan 4, 2021 at 13:38
  • $\begingroup$ @RiverLi $\;\lim\limits_{n\to \infty}\frac{n-1}n=1.\;$ :-) $\endgroup$ Jan 4, 2021 at 13:46
  • $\begingroup$ But $x$ depends on $n$, $\lim_{n\to \infty} x^{nx} = \lim_{n\to \infty} (\frac{n-1}{n})^{n-1} = \mathrm{e}^{-1}$. $\endgroup$
    – River Li
    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= $$\int_{0}^{1}1dx$$


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


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