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I have some trivial confusion.

In the following case, is the limit $a$? or is that the limit does not exist? I am just trying to be too careful not to do something wrong. Specially when infinity is so deceitful.

$$\lim_{n\rightarrow\infty}\sum^{n}_{i=1}\frac{a}{n}=a?/\infty??$$

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    $\begingroup$ $\sum_i \frac{a}{n} = a \sum_i \frac{1}{n} = a \frac{1}{n} \sum_i 1$. This is about the series, not $a$. What is the value of $\sum_i 1$? $\endgroup$ – copper.hat Feb 27 '13 at 23:53
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    $\begingroup$ No. $\sum_{i=1}^n 1 = 1+...+1$ ($n$ times). Hence $\sum_{i=1}^n 1 = n$. And $\frac{1}{n} \sum_{i=1}^n = 1$. $\endgroup$ – copper.hat Feb 28 '13 at 0:01
  • $\begingroup$ @user1709828 It's not valid to split it up as $(\lim_{n\to\infty} \sum_{i=1}^n a) \cdot (\lim_{n\to\infty} \sum_{i=1}^n \frac{1}{n})$, but I don't see why you would want to do that anyway. Just evaluate the thing you're taking the limit of. $\endgroup$ – Trevor Wilson Feb 28 '13 at 0:04
  • $\begingroup$ A you sure it is not $a \over i$ in the sum? $\endgroup$ – Maesumi Feb 28 '13 at 0:37
  • $\begingroup$ @Maesumi then it would not confuse me. $\endgroup$ – user45099 Feb 28 '13 at 6:47
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The other answers all factor out the $a$ for some reason, but it's easier just to notice that $$ \sum_{i=1}^n \frac{a}{n} = \underbrace{\frac{a}{n} + \cdots + \frac{a}{n}}_{\text{$n$ times}} = n \cdot \frac{a}{n} = a,$$ so you are taking the limit of a constant sequence. What is $\lim_{n \to \infty}a$?

By the way, I don't know if this is causing your confusion, but $\lim_{n \to \infty} \sum_{i=1}^n \frac{a}{n}$ is not the same as $\lim_{n \to \infty} \sum_{i=1}^n \frac{a}{i}$. The latter limit is by definition the sum of the series $\sum_{i=1}^\infty \frac{a}{i}$, which diverges (if $a$ is nonzero.)

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Your sum is: $$ \lim_{n \rightarrow \infty} \sum_{1 \le k \le n} \frac{a}{n} = a \lim_{n \rightarrow \infty} \sum_{1 \le k \le n} \frac{1}{n} = a \lim_{n \rightarrow \infty} 1 = a $$

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