Evaluate $\lim _{n\to\infty}\left[\frac{1}{n+1}+\frac{1}{n+2}+\cdots \frac{1}{n+n}\right]$

Question

Evaluate $\lim _{n\to\infty} \left[\frac{1}{n+1}+\frac{1}{n+2}+\ldots \frac{1}{n+n}\right]$

My solution goes like this:

We consider $n$ sequences, $x_1=\frac{1}{n+1}, x_2=\frac{1}{n+2},\ldots , x_n=\frac{1}{2n}.$ Now, $\lim_{n\to\infty}x_k=0$ , $\forall k\in \{1,2,\ldots ,n\}$ and thus,$\lim _{n\to\infty} \left[\frac{1}{n+1}+\frac{1}{n+2}+\ldots \frac{1}{n+n}\right]=\lim_{n\to \infty} [x_1+\ldots +x_n]=0.$

Is the above solution correct? If not, where is it going wrong?

EDIT 1: As suggested by Bernkastel and others, the reason for this solution to be incorrect is justified. But, this link : For each positive integer $n$, let $x_n=1/(n+1)+1/(n+2)+\cdots+1/(2n)$. Prove that the sequence $(x_n)$ converges. does not answer my question as I wanted to know, why my solution is not valid and was not interested in any alternative solution of this problem. This was the intended while positing this initially.

Furthermore, is there any alternative solution which, does not involve Riemann sums or Riemann integrals? As, I still haven't learnt them. To be specific, I only know, the fundamental properties and the elementary definition of definite integrals. The "Riemann integral" is completely unknown to me, as of yet.

EDIT 2: I have added my own answer, which clarifies all these issues to the best of my understanding. Many thanks, To all the community members for their benignant co-operations in all their travails !

Answer 1

The method in the OP is not a valid solution. This is because the theorem utilized in the OP is :

If $x_n$ and $y_n$ are two two convergent series such that $\lim_{n\to\infty} x_n=x$ and $lim_{n\to\infty} y_n=y$ then $\lim_{n\to\infty}(x_n+y_n)=x+y.$

But notice, that the above theorem is valid or rather, can be extended for a finite number of sequences but in the posed problem, the number of sequences considered in the solution portion is infinite as, $x_1,x_2,\cdots,x_n$ are the sequences considered and $n$ is infinite, not a finite quantity.

The real solution can be given in the following way:

Given, $\lim_{n\to\infty}[\frac{1}{n+1}+\cdots +\frac{1}{n+n}]=\lim_{n\to\infty} \frac{1}{n}[\frac{1}{1+\frac 1n}+\cdots +\frac{1}{1+\frac nn }].$ We consider, $f(x)=\frac{1}{1+x}.$ Now, we can say, $\lim_{n\to\infty} \frac{1}{n}[\frac{1}{1+\frac 1n}+\cdots +\frac{1}{1+\frac nn }]=\lim_{n\to\infty} \frac{1}{n}[f(\frac 1n)+\cdots +f(\frac nn)].$ Thus, the thing going on here, is, that if there is an interval $[0,1]$ we can break up this interval into smaller intervals of $[0,\frac 1n],[\frac1n,\frac 2n],\cdots ,[\frac{n-1}n,\frac nn].$ Now, we know that $\int_a^bf(x)dx=\lim_{n\to\infty}\frac {b-a}{n}(f(a)+f(a+\frac {b-a}{n})+f(a+2\frac {b-a}{n})+\cdots +f(a+n\frac{b-a}{n})),$ if $f$ is continuous and differentiable on the interval $[a,b].$ We use this theorem, in our case as well. This is a standard strategy. Now, in our case, $a=0,b=1$ and all the criterions of continuity and differentiability are satisfied by $f$ and hence, $L=\lim_{n\to\infty} \frac{1}{n}[\frac{1}{1+\frac 1n}+\cdots +\frac{1}{1+\frac nn }]=\lim_{n\to\infty} \frac{1}{n}[f(\frac 1n)+\cdots +f(\frac nn)]=\int_0^1 f(x)dx=\int_0^1\frac{1}{1+x}dx=[\log(1+x)+c]^1_0,$ where $c$ is an arbitary constant of integration. Thus, $L=\log 2.$

Thus, $\lim_{n\to\infty}[\frac{1}{n+1}+\cdots+\frac{1}{n+n}]=\log 2$

The above solution, uses only fundamental results and typically it does not involve Riemann sums in the way, but uses a basic approach which is actually the essence of definite integrals. The solution, so presented do not refer to Riemann integrals or sums directly.

Answer 2

$\begin{align}&\lim _{n\to\infty} [\frac{1}{n+1}+\frac{1}{n+2}+\cdots \frac{1}{n+n}]\\&=\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{n+k}\\&=\lim_{n\to\infty}\sum_{k=1}^{n}f_k\end{align}$

where $f_k=\frac{1}{k+n}$

$f_k\to 0$ for all $k\in\{1, 2,\ldots n\}$

Does this implies $\sum_{k=1}^{n}f_k\to 0$?

Convergence of a series (or sequence) doesn't depends on finitely many terms!

Answer 3

No, this is wrong. Note that ${1 \over n+k} \geq {1\over 2n}$ for $1\leq k\leq n$ so that we have for all $n$, $${1\over n+1}+ ....+ {1\over 2n} \geq n\times {1\over 2n} = {1\over 2}$$ This is an example where you can't invert sums and limits.

You need to convert your expression into a Riemann sum to compute the limit.

Edit: at some point, the OP asked for the computation of the limit without using Riemann sums. After some thought, here is my solution.

First, it is an elementary result, explained in this other post, to show that the following sequence is convergent. $$ \sum_{k=1}^n {1 \over k} - \ln(n)\longrightarrow \gamma. $$ Here $\gamma$ is some constant, actually Euler's constant, whose value we won't need. Let us denote by $\varepsilon(n) $ the difference between the previous sequence and its limit. We are almost done. $$ \sum_{k=1}^n {1\over n+k} = \sum_{k=1}^{2n} {1\over k} - \sum_{k=1}^n {1\over k} = \ln(2n)+\gamma -\ln(n)-\gamma + \varepsilon(2n)-\varepsilon(n) = \ln(2)+ \varepsilon(2n)-\varepsilon(n). $$ The limit is $\ln(2)$ as expected.

Answer 4

As others have said, the problem with the initial argument is the misuse of the theorem that the limit of a sum of a fixed and finite number of sequences is the sum of the limits of those sequences. However, the number of sequences is infinite or increasing, depending on how you look at it.

Here is a proof of the convergence and the evaluation of the limit using only pre-calculus (sums and limits, but no integration or differentiation).

---

Convergence

Note that the sequence is increasing: $$ \begin{align} \sum_{k=m+1}^{2m}\frac1k-\sum_{k=m}^{2m-2}\frac1k &=\frac1{2m}+\frac1{2m-1}-\frac1m\tag{1a}\\ &=\frac1{2m-1}-\frac1{2m}\tag{1b}\\[6pt] &\gt0\tag{1c} \end{align} $$ Explanation:
$\text{(1a):}$ cancel the common terms in the series
$\text{(1b):}$ simplify
$\text{(1c):}$ $\frac1{2m-1}\ge\frac1{2m}$

Furthermore, the sequence is bounded by $1$: $$ \begin{align} \sum_{k=n+1}^{2n}\frac1k &\lt\sum_{k=n+1}^{2n}\frac1n\tag{2a}\\[3pt] &=1\tag{2b} \end{align} $$ Explanation:
$\text{(2a):}$ each term in the sum is less than $\frac1n$
$\text{(2b):}$ there are $n$ terms in the sum

Thus, $\sum\limits_{k=n+1}^{2n}\frac1k$ is an increasing sequence that is bounded above by $1$. Therefore, the sequence converges to a limit not exceeding $1$.

---

Limit

Using Bernoulli's Inequality, Telescoping Products, and the Squeeze Theorem, we can compute the limit.

Lemma: For all $x\in\mathbb{R}$, $$ e^x\ge1+x\tag3 $$ Proof: For $x\lt-1$, $(3)$ is immediate since the left side is positive and the right side is negative. So we will assume that $x\ge-1$.

Define $$ e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n\tag4 $$ For $n\ge1$, Bernoulli's Inequality says that $$ \left(1+\frac xn\right)^n\ge1+x\tag5 $$ Taking the limit of $(5)$ as $n\to\infty$ and applying $(4)$ yields $(3)$.

$\large\square$

Substituting $x\mapsto-x$ in $(3)$ and taking reciprocals yields, for $x\lt1$, $$ e^x\le\frac1{1-x}\tag6 $$ Applying $(3)$ gives $$ \begin{align} e^{\frac1{n+1}+\frac1{n+2}+\cdots+\frac1{2n}} &\ge\frac{n+2}{n+1}\frac{n+3}{n+2}\cdots\frac{2n+1}{2n}\tag{7a}\\[6pt] &=\frac{2n+1}{n+1}\tag{7b}\\[4 pt] &=2-\frac1{n+1}\tag{7c} \end{align} $$ Explanation:
$\text{(7a):}$ apply $(3)$
$\text{(7b):}$ telescoping product
$\text{(7c):}$ simplify

Applying $(6)$ gives $$ \begin{align} e^{\frac1{n+1}+\frac1{n+2}+\cdots+\frac1{2n}} &\le\frac{n+1}n\frac{n+2}{n+1}\cdots\frac{2n}{2n-1}\tag{8a}\\[6pt] &=\frac{2n}n\tag{8b}\\[9 pt] &=2\tag{8c} \end{align} $$ Explanation:
$\text{(8a):}$ apply $(6)$
$\text{(8b):}$ telescoping product
$\text{(8c):}$ simplify

Taking logs of $(7)$ and $(8)$ yields $$ \log\left(2-\frac1{n+1}\right)\le\frac1{n+1}+\frac1{n+2}+\cdots+\frac1{2n}\le\log(2)\tag9 $$ Taking the limit of $(9)$ as $n\to\infty$, the Squeeze Theorem gives $$ \lim_{n\to\infty}\sum_{k=n+1}^{2n}\frac1k=\log(2)\tag{10} $$