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I came across a question in my High School Calculus textbook:

Find $\lim_{n \to \infty} a_n$ where $a_n = \frac{1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2} + ... + \frac{n}{n^2}$

My approach was to simply distribute the limit to each term in the partial sum. I believed this would be legitimate according to the limit laws. Once I evaluated the individual limits, the limit of the partial sum came out to equal $0$:

$$\begin{aligned} \lim_{n \to \infty} a_n &= \lim_{n \to \infty}(\frac{1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2} + ... + \frac{n}{n^2}) \\ &= \lim_{n \to \infty}(\frac{1}{n^2}) + \lim_{n \to \infty}(\frac{2}{n^2}) + \lim_{n \to \infty}(\frac{3}{n^2}) + ... + \lim_{n \to \infty}(\frac{n}{n^2}) \\ &= 0+0+0+...+0 \\ &=0 \end{aligned}$$

However, the textbook first combined the fractions and simplified it, only to find that the limit of the partial sum equals $1/2$:

$$\begin{aligned} a_n &= \frac{1+2+3+...+n}{n^2} \\ &= \frac{\frac{n(n+1)}{2}}{n^2} \\ &= \frac{1}{2}(\frac{n+1}{n}) \\ &= \frac{1}{2}(1+\frac{1}{n}) \end{aligned}$$

$$\therefore \lim_{n \to \infty}a_n = \frac{1}{2}$$

Both solutions seemed to be legitimate. However, after pondering on the question for a bit, I starting to think that maybe the limit laws don't apply to partial sums... As $n\rightarrow\infty$, the number of terms also approaches infinity, so then distributing the limit an "infinite" number of times might be wrong, considering how weird infinity is. But I'm not so convinced about that.

Is there some sort of restriction on how the limit can be distributed? Is there something I'm missing that completely invalidates my approach to the problem?

scrap
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    Another example: $1 = \frac 1n + \frac 1n + \ldots + \frac 1n \to 0 + 0 + \ldots + 0 = 0$ – where is the fault? – Martin R May 10 '20 at 21:04
  • See also https://math.stackexchange.com/q/3164457/42969. – Martin R May 10 '20 at 21:06
  • I was really close to finishing up a long analytical answer when this question closed—pretty annoying. – Brian Tung May 10 '20 at 21:19
  • @BrianTung Oh shoot, I'm very sorry about that. Is there any way you can share your answer to me? I would really appreciate it. – scrap May 10 '20 at 21:23
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    @ilsarsari: In this case, it's fairly simple; I moved it to the question this question is marked as a duplicate of. :-) – Brian Tung May 10 '20 at 21:26

2 Answers2

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The actual rule is this: If $a_n \to A$ and $b_n \to B$, then $a_n + b_n \to A + B$. The big technicality is that both limits must exist and be finite. You cannot generally do "limit of a sum is the sum of limits" if you do not know if all limits exist.

The solution in the text is the preferred way to solve this, as it manipulates the sum in a way that's familiar using the triangle numbers.

Sean Roberson
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  • Thank you for your answer! I agree that manipulating the sum first is usually preferred. However, considering how simple that series is, we do know that all the limits exist and are finite, right? – scrap May 10 '20 at 20:54
  • Well - no. When looking at the limit of a sum where the number of terms is possibly infinite, we need to be extremely careful. We can't just pass the limit through the sum. – Sean Roberson May 10 '20 at 20:55
  • Is there a reason for this? What are the implications of an infinite number of terms? I'm just trying to wrap my head around this, thanks! – scrap May 10 '20 at 20:59
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    The real problem is that the number of summands depends on $n$ in the first solution, not that the individual limits do not exist. – Martin R May 10 '20 at 21:07
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Perhaps too trivial.

Example:

1) $1= n(1/n)= $

$1/n +1/n +...+1/n,$ $n$ terms;

Take the limit $n \rightarrow \infty$.

2) $1000/n=$

$1/n +1/n+ ..1/n$, $1000$ terms.

Take the limit $n\rightarrow \infty$.

Peter Szilas
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