What Went Wrong?

First of all, let's check whether the value of the limit obtained is correct.

In the sum of n, n², or n³ wiki, we saw that $1 + 2 +3 + \cdots + n = \frac{n(n+1)}{2}$ . We can substitute this in the problem.

$\begin{aligned} L & = \lim _{n \to \infty} \frac{1 + 2+ 3 + \cdots + n}{n^{2}} \\ & = \lim _{n \to \infty} \frac{n(n+1)}{2n^2} \\ & = \lim _{n \to \infty} \frac{n+1}{2n} \\ & = \lim _{n \to \infty} \frac{1 + \frac{1}{n}}{2} \end{aligned}$

When $n \to \infty$ , we see that $\frac{1}{n} \to 0$ . Therefore $L = \frac{1}{2}$ . The answer does not match with the answer obtained in the problem, so at least one of the steps in the problem is not valid.

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• There is nothing wrong in Step 1 . The limits exists, and we just found its value.

• There is nothing wrong in Step 2 either. Both the numerator and the denominator tend to positive infinity. The limit is of the indeterminate form $\frac{\infty}{\infty}$ .

• However Step 3 is wrong. Since L'hôpital's rule is generally defined for functions whose domain is real numbers and the function in the problem has domain of natural numbers, we cannot apply it directly. Moreover, while trying to apply L'hôpital's rule in Step 3, it is assumed that $\frac{d}{dn} (1 + 2 +3 + \cdots + n ) =1$ , which is incorrect.

Instead, let us define a function $f \colon \mathbb{R} \to \mathbb{R}$ such that $f(x) = \frac{x(x+1)}{2x^2}$ . We can now apply L'hôpital's rule.

$\begin{aligned} \lim_{x \to \infty} f(x) & = \lim_{x \to \infty} \frac{x(x+1)}{2x^2} \\ & = \lim_{x \to \infty} \frac{2x + 1}{4x} \\ & = \lim_{x \to \infty} \frac{2 + \frac{1}{x}}{4} \\ \end{aligned}$

When $x \to \infty$ , we see that $\frac{1}{x} \to 0$ . Therefore $\lim _{x \to \infty} = \frac{2}{4} = \frac{1}{2}$ .

Note that for every natural number $n$ , $f(n)$ is identical to the given expression in the problem: $\frac{1+ 2 + 3 + \cdots + n}{n^{2}}$ . As $x$ tends to $+ \infty$ , we see that $f(x)$ tends to $\frac{1}{2}$ . So as $n$ tends to $+\infty$ , $f(n)$ also tends to $\frac{1}{2}$ . Therefore $L = \frac 12$ .

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Alternatively, we can apply the Stolz–Cesàro theorem , which is a discrete version of L'hôpital's rule. It states that

• if $\{a_n\}_{n \geq 1}$ and $\{b_n\}_{n \geq 1}$ are sequences of real numbers, and

• if $\{b_n\}_{n \geq 1}$ is a strictly monotonic divergent sequence, and

• if $\displaystyle \lim _{n \to \infty} \frac{a_n}{b_n}$ is finite and is equal to $L$ , then

$\lim _{n \to \infty} \frac{a_n}{b_n} = L = \lim _{n \to \infty} \frac{a_{n+1} - a_n } {b_{n+1} - b_n }$

In the given problem, $a_n = 1 + 2 + 3 + \cdots + n$ and $b_n = n^{2}$ . Note that $\{b_n\} = n^2$ is a strictly increasing and divergent sequence.

$\begin{aligned} L & = \lim _{n \to \infty} \frac{1 + 2+ 3 + \cdots + n} {n^{2}} \\ & = \lim _{n \to \infty} \frac{(1 + 2 + 3 + \cdots + n + (n + 1)) - (1 + 2 + 3 + \cdots + n) }{(n+1)^2 - n^{2}} \\ & = \lim _{n \to \infty} \frac{n+1}{2n+1} \\ & = \lim _{n \to \infty} \frac{1 + \frac{1}{n}}{2 + \frac{1}{n} } \end{aligned}$

When $n \to \infty$ , we see that $\frac{1}{n} \to 0$ . Therefore $L = \frac{1}{2}$ .

There is nothing wrong with step 1 and 2.

In step 3, applying l'Hôpital's rule as $\displaystyle \lim_{n \to \infty} \frac{\frac{d}{dx} (1+2+3+...+n)}{\frac{d}{dx} n^2}$ is wrong. It should be as follows:

$\begin{aligned} \mathfrak{L} & = \lim_{n \to \infty} \frac{\color{#3D99F6}{1+2+3+...n}}{n^2} \\ & = \lim_{n \to \infty} \frac{\color{#3D99F6}{n(n+1)}}{\color{#3D99F6}{2}n^2} \\ & = \lim_{n \to \infty} \frac{n^2+n}{2n^2} \quad \quad \small \color{#3D99F6}{\text{Differentiate up and down.}} \\ & = \lim_{n \to \infty} \frac{2n+1}{4n} \quad \quad \small \color{#3D99F6}{\text{Differentiate up and down again.}} \\ & = \lim_{n \to \infty} \frac{2}{4} \\ & = \frac{1}{2} \end{aligned}$

Limit can also be obtained as follows:

$\begin{aligned} \mathfrak{L} & = \lim_{n \to \infty} \frac{n^2+n}{2n^2} = \lim_{n \to \infty} \left( \frac{1}{2} + \frac{1}{2n} \right) = \frac{1}{2} \end{aligned}$