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We have $$\dfrac{1+2+3+...+ \space n}{n^2}$$

What is the limit of this function as $n \rightarrow \infty$?

My idea:

$$\dfrac{1+2+3+...+ \space n}{n^2} = \dfrac{1}{n^2} + \dfrac{2}{n^2} + ... + \dfrac{n}{n^2} = 0$$

Is this correct?

mdp
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LimitsMate
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  • When you break it down in terms, each individual term goes to zero, but their number goes to infinity. Thus, in "limit", you get infinitely many zeroes, or in other words $\infty \cdot 0$... But this can be anything.... – N. S. May 14 '13 at 16:09

4 Answers4

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HINT:

Such application of limit (esp. to $\infty$) to individual summand is applicable only when the number of summand is finite.

$$\dfrac{1+2+3+...+ \space n}{n^2} =\frac{\frac{n(n+1)}2}{n^2}=\frac12\cdot\left(1+\frac1n\right)$$

lab bhattacharjee
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  • So it is $\dfrac{1}{2}$. I however don't understand why my method is wrong, and also your first step eludes me – LimitsMate May 14 '13 at 15:23
  • You seem to be trying to use the sum rule for sequences - this only works when the number of terms in the sum is constant, but in this case it grows with $n$. The first step in this answer is the result $\sum_{i=1}^ni=\frac{1}{2}n(n+1)$. – mdp May 14 '13 at 15:26
  • @LimitsMate, please find the explanation in the answer – lab bhattacharjee May 14 '13 at 15:26
  • @LimitsMate, For example $$\lim_{n\to\infty}\left(1+\frac1n\right)^n\ne 1$$ – lab bhattacharjee May 14 '13 at 15:30
  • @labbhattacharjee Coincidentally I know that's $e$. Wow, that's the best counterexample you could have given lol – LimitsMate May 14 '13 at 15:31
  • @LimitsMate, could I make the idea clear? – lab bhattacharjee May 14 '13 at 15:35
  • @labbhattacharjee Yes, thanks, although the idea is hard for me to fathom. Is there perhaps a formal proof to your first statement? – LimitsMate May 14 '13 at 15:36
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    @labbhattacharjee: I'm a bit puzzled about the (added) first sentence and the interversion limit/sum (the way it is phrased, I'm not sure it is correct -- see for instance the theorems for series of functions when you have uniform convergence — or even that I understand the statement —as "$n$" is used as a summation indice (I assume) and upper bound (ditto)) – Clement C. May 14 '13 at 15:38
  • @ClementC., how about the re-phrased version? – lab bhattacharjee May 14 '13 at 16:02
  • Seems fine to me :) – Clement C. May 14 '13 at 16:40
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Another method, using Riemann sums applied to $f:x\mapsto x$: $$ \frac{\sum_{k=1}^n k }{n^2} = \frac{1}{n}\sum_{k=1}^n \frac{k }{n} \xrightarrow[n\to\infty]{} \int_{0}^{1} xdx $$

As for your question, each term goes to $0$, but the number of terms grows; you are not in the case where you have the sum of "constantly many terms, each of them converging".

Clement C.
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Recall that $$1 + 2 + \cdots + n =\sum_{i = 1}^n i = \frac{n(n+1)}{2}$$

So you want $$\lim_{n\to \infty}\dfrac{n(n+1)}{2n^2} = \lim_{n\to\infty}\frac{n^2+ n}{2n^2} = \lim_{n\to \infty} \left(\frac 12 + \frac 1{2n}\right)$$

amWhy
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Your method is wrong. Consider the first values of the expression:

  • for $n=1$ you get $1$;
  • for $n=2$ you get $1/4+2/4=3/4$;
  • for $n=3$ yoy get $1/9+2/9+3/9=6/9=2/3$.

One could make the conjecture that these terms are always bigger than $1/2$, which can be proved by induction. Let $$ f(n)=\frac{1}{n^2}\sum_{i=1}^n i $$ We have $f(1)=1>1/2$. Suppose the assertion holds for $f(n)$ and consider \begin{align} f(n+1)&=\frac{1}{(n+1)^2}\sum_{i=1}^{n+1} i\\ &=\biggl(\frac{1}{(n+1)^2}\sum_{i=1}^{n} i\biggr)+\frac{n+1}{(n+1)^2}\\ &=\frac{n^2}{(n+1)^2}\biggl(\frac{1}{n^2}\sum_{i=1}^n i\biggr)+\frac{1}{n+1}\\ \text{(by induction hypothesis)}\qquad&>\frac{1}{2}\frac{n^2}{(n+1)^2}+\frac{1}{n+1}\\ &=\frac{1}{2}\frac{n^2+2n+2}{(n+1)^2}\\ &=\frac{1}{2}\biggl(\frac{n^2+2n+1}{(n+1)^2}+\frac{1}{(n+1)^2}\biggr)\\ &=\frac{1}{2}\biggl(1+\frac{1}{(n+1)^2}\biggr)\\ &>\frac{1}{2} \end{align} Therefore your conclusion that the limit is $0$ cannot be true.

egreg
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