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Evaluate the limit $$\lim_{n\to\infty} \left(\frac{1}{n^2}+\frac{2}{n^2}+...+\frac{n-1}{n^2}\right) .$$

My work:

I started by computing the first 8 terms of the sequence $x_n$ ($0, 0.25, 0.222, 0.1875, 0.16, 0.139, 0.122, 0.1094$). From this, I determined that the sequence $x_n$ monotonically decreases to zero as n approaches infinity. which satisfies my first test for series convergance, if $\sum_{n=2}^\infty x_n$ converges, then $\lim_{n\to\infty}x_n=0$.

Next, I rearranged the equation in an attempt to perform a comparison test. I re-wrote the equation as $\sum_{n=2}^\infty (\frac1{n}-\frac1{n^2})$. This was to no avail as the only series I could think to compare it to was $\frac1n$ which is always greater than the original series and is divergant, which does not prove convergance to a limit.

The ratio test concluded with $\lim_{n\to\infty} \frac{x_{n+1}}{x_n}$ being equal to 1, which is also inconclusive (I can show my work if need be, but that would be a little tedious). I never ran the root test, but I doubt that this would be any help in this case.

I see no other way to compute the limit, so any help would be appreciated!!

Travis Willse
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Kevin McDonough
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4 Answers4

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Hint: Use the formula $$\sum_{k=1}^n k=\frac{n(n+1)}2.$$

So, $$\lim_{n\to\infty}\frac 1{n^2}+\frac 2{n^2}+\cdots+\frac{n-1}{n^2}=\lim_{n\to\infty}\frac{n(n-1)}{2n^2}.$$

Can you get it from here?

Tim Raczkowski
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  • Could you be more specific? Use it for a comparison test? – Kevin McDonough Oct 18 '15 at 17:00
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    It is not instantly obvious to me, but it sounds like an excellent starting point. I'm going to go through some old notes and try to piece together the solution. Thank you for the response! I will be sure to credit you when I find a solution! – Kevin McDonough Oct 18 '15 at 17:06
  • Figured it out! Thank you so much! – Kevin McDonough Oct 18 '15 at 17:20
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    Happy to help. :) – Tim Raczkowski Oct 18 '15 at 17:21
  • Actually I'm still a little confused. I used your equation above and found that the limit should be equal to $\frac12$. However, when I add up the first 6 terms in the sequence, they sum to a number greater than this limit. Assuming that my answer is correct, why is this the case? – Kevin McDonough Oct 18 '15 at 17:31
  • Not sure. Something to keep in mind: as the number of terms increases, so does the denominator. – Tim Raczkowski Oct 18 '15 at 17:34
  • @KevinMcDonough Your answer isn't correct :). You're trying to see the question as an infinite series when it really isn't. Compare (without summing) the values of the expression in the limit for $n=2$ ($\tfrac14$), $n=3$ ($\tfrac19 + \tfrac29$), $n=4$ ($\tfrac1{16} + \tfrac2{16} + \tfrac3{16}$). These are not partial sums of a common series, so it makes no sense to talk about $\sum x_n$ when the question is actually about $\lim x_n$. – Erick Wong Oct 18 '15 at 17:34
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Hint: Recall the definition of integration.(Riemann Integration)

$$\lim_{n\rightarrow\infty}\sum_{k=1}^{n-1}\frac{k}{n^2}$$ $$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^{n-1}\frac{k}{n}$$ $$\lim_{n\rightarrow\infty}\frac{1}{n} \sum_{k=1}^{n-1} f\left(\frac{x}{n}\right)=\int_{0}^{1}x \; dx$$ $$\frac{x^2}{2}\Big|_{0}^{1}=\frac{1}{2}$$

ReverseFlowControl
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2

This is a a "left-handed" Riemann sum with $n$ terms approximating $\int_0^1 x dx$ with a step size of $\frac{1}{n}$, so the limit is the value of this integral, which is $\frac{1}{2}$.

hunter
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Hint We can rearrange the argument of the limit as $$\frac{1}{n} \sum_{k = 0}^{n - 1} \frac{k}{n}.$$ Can you interpret this as a Riemann sum for some function and hence realize the given limit as an (easy-to-evaluate) definite integral?

Travis Willse
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