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$$ \lim_{n\to\infty}\left(\frac{1}{n^2}+\frac{2}{n^2}+\frac{3}{n^2}+\ldots+\frac{n}{n^2}\right)$$

$$ = \lim_{n\to\infty}\left(\frac{1+2+3+\ldots+n}{n^2} \right)$$

$$ = \lim_{n\to\infty}\left(\frac{n+1}{2n} \right)=\frac {1}{2} $$

I have a question here, Can we use The limit rule like $$ \lim_{n\to\infty}[f(x)+g(x)]=\lim_{n\to\infty}f(x)+\lim_{n\to\infty}g(x)$$ to conclude:$$ \lim_{n\to\infty}\left(\frac{1}{n^2}+\frac{2}{n^2}+\frac{3}{n^2}+\ldots+\frac{n}{n^2}\right)$$$$=\lim_{n\to\infty}\frac{1}{n^2}+\lim_{n\to\infty}\frac{2}{n^2}+\lim_{n\to\infty}\frac{3}{n^2}+\ldots+\lim_{n\to\infty}\frac{n}{n^2}$$$$=0+0+\ldots+0=0$$ which is absurd. But I can't figure out where the error is. Please help me out.

user
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Andrew Lion
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  • Thanks, can you post any link? @geetha290krm – Andrew Lion Jun 26 '23 at 08:43
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    Related https://math.stackexchange.com/q/4720684/505767 – user Jun 26 '23 at 09:17
  • I know what @geetha290krm implies with stating "very big NO" , though it is Equally true to state "very big YES" : We will have $0+0+0+0+0+0 \cdots = \infty \times 0$ which is "not automatically" $0$ , It implies that we have to try with other ways to get that limit Correctly !! – Prem Jun 26 '23 at 10:31

1 Answers1

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It is indeed true that if $(a_n)_n$ and $(b_n)_n$ are converging sequence, then so is $(a_n+b_n)$ and its limit is $\lim_{n\to\infty}a_n+\lim_{n\to\infty}b_n$. This extends to a finite number of sequences.

However, the thing get worst when we do not have a fixed number of sequences. The Riemann sums in the opening post can be written as $$ \sum_{j=1}^\infty a_{n,j}, a_{n,j}=jn^{-2}\mathbf{1}_{j\leqslant n}. $$ It is indeed true that for each fixed $j$, $\lim_{n\to\infty}a_{n,j}=0$. However, for a fixed $n$, the sum over $j$ is already a limit, namely, $\lim_{N\to\infty}\sum_{j=1}^Na_{n,j}$ and in general, switching limit may be problematic. For example, if $b_{n,N}=\mathbf{1}_{n\leqslant N}$, then $\lim_{n\to\infty}\lim_{N\to\infty}b_{n,N}=1$ but $\lim_{N\to\infty}\lim_{n\to\infty}b_{n,N}=0$.

Davide Giraudo
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