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Now, I know this to be correct:

$$\begin{align*} \lim_{n \rightarrow\infty} \left(\frac 1{n^2}+\frac 2{n^2}+\ldots+\frac n{n^2}\right)&=\lim_{n \rightarrow\infty} \left[\frac 1{n^2} \left(\frac n2\right)(1+n)\right]\\ &=\lim_{n \rightarrow\infty} \frac {1+n}{2n}\\ &=\frac 12\;. \end{align*}$$

But what is wrong with the following reasoning?

$$\begin{align*} \lim_{n \rightarrow\infty} \left(\frac 1{n^2}+\frac 2{n^2}+\ldots+\frac n{n^2}\right)&=\lim_{n \rightarrow\infty} \frac 1{n^2} + \displaystyle \lim_{n \rightarrow\infty} \frac 2{n^2} +...+ \displaystyle \lim_{n \rightarrow\infty} \frac n{n^2}\\\\ &=0+0+\ldots+0\\\\ &=0\;? \end{align*}$$

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  • $\begingroup$ See also here. $\endgroup$ – Martin Sleziak May 30 '12 at 11:01
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    $\begingroup$ If you have an awful lot of little things, they can add up to a big thing. $\endgroup$ – André Nicolas May 30 '12 at 13:38
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The error is that in your reasoning $$ \lim_{n \rightarrow\infty} (\frac 1{n^2}+\frac 2{n^2}+...+\frac n{n^2}) $$

$$=\lim_{n \rightarrow\infty} \frac 1{n^2} + \displaystyle \lim_{n \rightarrow\infty} \frac 2{n^2} +...+ \displaystyle \lim_{n \rightarrow\infty} \frac n{n^2} $$

you have a finite addition instead of the infinite. Now, to be precise, formal and serious - if you have the series $\sum\limits_{k=1}^n a_k(n)$ then $$ \lim\limits_n\sum\limits_{k=1}^n a_k(n)\neq \sum\limits_{k=1}^n \lim\limits_n a_k(n), $$ the most simple reason being that the LHS does not depend on $n$ while RHS does - which is a consequence of taking the limit under the sum, where the sum bounds themselves depend on $n$.

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  • $\begingroup$ I think I almost understand, but I don't get why the sum rule for limits doesn't apply here. The Sum Rule being: If the limits of $a_n$ and $b_n$ are $l$ and $m$ respectively, then the limit of $a_n + b_n$ equals $l+m$. Help? $\endgroup$ – Ryan G May 30 '12 at 13:20
  • $\begingroup$ @Ryan: that is always true for two sequences $a_n,b_n$ - hence it is true for any fixed number of sequences $a^1_n,a^2_n,\dots,a^m_n$ but not in the case the number of sequences depends on $n$ by itself. $\endgroup$ – Ilya May 30 '12 at 13:23
  • $\begingroup$ Oh, thanks! I just realised that you already pointed this out in your original answer, hee. $\endgroup$ – Ryan G May 30 '12 at 13:33
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Look at $$\frac1{n^2}+\frac2{n^2}+\ldots+\frac{n}{n^2}$$ for the first few values of $n$:

$$\begin{array}{c} \frac11\\ \frac14&+&\frac24\\ \frac19&+&\frac29&+&\frac39\\ \frac1{16}&+&\frac2{16}&+&\frac3{16}&+&\frac4{16}\\ \vdots&&\vdots&&\vdots&&\vdots&&\ddots \end{array}$$

It’s true that each column is converging to $0$, but the number of columns is increasing at the same time, and you don’t know until you work it out which effect is going to ‘win’. In this case neither wins: they almost exactly balance out, with an overall limit of $1/2$.

It’s somewhat analogous to $0\cdot\infty$ indeterminate forms, limits of the form $\lim\limits_{x\to\infty}f(x)g(x)$ where $\lim\limits_{x\to\infty}f(x)=0$ and $\lim\limits_{x\to\infty}g(x)=\infty$: here $f$ can win, making the limit $0$, or $g$ can win, making it $\infty$, or they can roughly balance out, making it some positive real number.

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    $\begingroup$ Nice array. ${}{}{}$ $\endgroup$ – davidlowryduda May 30 '12 at 10:42

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