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I have tried to find this limit using two different methods which yield two different answers. I need clarification as to which method gives the correct result. The first method yields, $$\lim_{n \to \infty} \frac{1 + 2 + \dots + n}{n^2}=\lim_{n \to \infty} \frac {\frac{1}{2} n(n+1)}{n^2} = \lim_{n \to \infty} \frac{\frac{1}{2}(1+\frac{1}{n})}{1} = \frac{1}{2} $$ However, the second method yields, $$\lim_{n \to \infty} \frac{1 + 2 + \dots + n}{n^2}= \lim_{n \to \infty} \frac{1}{n^2} + \lim_{n \to \infty} \frac{2}{n^2} + \dots + \lim_{n \to \infty} \frac{n}{n^2} = 0 + 0+ \dots +0 = 0$$

Rodrigo de Azevedo
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user24334123
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    Second method is invalid – Sine of the Time Jan 15 '25 at 12:09
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    The limit of an infinite sum is generally not the infinite sum of the limits. – rafilou2003 Jan 15 '25 at 12:13
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    The second method is invalid. However, you can modify it slightly by using Stolz-Cesaro (the discrete version of L'Hopital) to obtain $\lim(1+\cdots + n) /n^2 = \lim(n+1) / ((n+1)^2-n^2) = \lim (n+1) / (2n+1) = 1/2$. – Integrand Jan 15 '25 at 12:13
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    @rafilou2003 There is no infinite sum here. – José Carlos Santos Jan 15 '25 at 12:14
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    @JoséCarlosSantos There is if you consider $\sum_{k=1}^n \frac{k}{n^2} = \sum_{k=1}^\infty a_{k,n}$ where $a_{k,n} = 0$ if $k > n$ and $a_{k,n} = \frac{k}{n^2}$ otherwise. In any case, we can agree there is no limit of a finite sum of terms – rafilou2003 Jan 15 '25 at 12:15
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    You can't use arithmetic of limits here, because it is not a sum of a constant number of sequences. The number of summands depends on $n$. – Mark Jan 15 '25 at 12:16
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    One way to convince yourself that the second approach has to be wrong is to consider the simpler example: $1 = n\cdot (1/n) = 1/n + 1/n + \ldots + 1/n$ and then take the limit to get the contradiction $1 = 0$. – Winther Jan 15 '25 at 12:29
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    The second way leads you to evaluating $\underbrace{0+0+\cdots + 0}_\infty = \infty\cdot 0,$ which is indeterminate. – CiaPan Jan 15 '25 at 12:34
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    One can express the sequence sums in the OP as a sequence of infinite sums $$s_n=\sum^n_{k=1}\frac{k}{n^2}=\sum^\infty_{k=1}c_{n,k}$$ by setting $$c_{n,k}=\frac{k}{n^2}\mathbf{1}{Z_n}(k)$$ where $Z_n={1,2,\ldots, n}$ and the the indicator function $\mathbf{1}{Z_n}(k)=1$ if $k\in Z_n$ and $\mathbf{1}{Z_n}(k)=0$ if $k\notin Z_n$. It is tempting to estimate $\lim_ns_n$ as $$\lim_ns_n=\lim_n\sum^\infty{k=1}c_{n,k}=\sum^\infty_{k=1}\lim_n c_{n,k}$$ This procedure, however is not valid in general for it may give contradictions such as $1/2=0$. (to be continued...) – Mittens Jan 15 '25 at 14:49
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    (continuation) Under some conditions, the limits of sequences of the form $$S_n=\sum^\infty_{k=1}a_{n,k}$$ can be shown to exists and can can be evaluates as $$\lim_n\sum^\infty_{k=1}a_{n,k}=\sum^\infty_{k=1}\lim_na_{n,k}$$ for example, if the following conditions hold: (1) for fixed $k$, $A_k=\lim_na_{n,k}$ exists and (2) there is a sequence $b_k$ such that $|a_{n,k}|\leq b_k$ and $\sum^\infty_{k=1}b_k$ converges. – Mittens Jan 15 '25 at 14:51

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If you had obtained

$$\text{exact }0+\text{exact }0+...+\text{exact }0$$

then indeed, the answer would have been $0$.

But what you have here is

$$\text{tending to }0+\text{tending to }0+...+\text{tending to }0$$

This is not zero. Instead, it is

$$(\text{tending to }0)\text{tending to infinite times}$$

which is an indeterminate form.

aarbee
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