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I'm studying for my calculus exam and I have the following limit:

$$\lim\limits_{n \to \infty} \left ( \frac{1}{\sqrt{n^3 +3}}+\frac{1}{\sqrt{n^3 +6}}+ \cdots +\frac{1}{\sqrt{n^3 +3n}} \right )$$

My solution is:

$$\begin{align*} &\lim\limits_{n\ \to \infty} \left ( \frac{1}{\sqrt{n^3 +3}}+\frac{1}{\sqrt{n^3 +6}}+ \cdots +\frac{1}{\sqrt{n^3 +3n}} \right )\\&= \lim_{n \to \infty}\left ( \frac{1}{\sqrt{n^3 +3}} \right ) + \lim_{n \to \infty}\left ( \frac{1}{\sqrt{n^3 + 6}} \right ) + \cdots + \lim_{n \to \infty} \left ( \frac{1}{\sqrt{n^3 +3n}} \right )\\ &=0 + 0 + \cdots + 0\\ &= 0 \end{align*}$$

It turned out to be suspiciously easy to solve.

Is this correct? If it isn't, what is wrong and how can I solve it correctly?

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You can't just split the terms up. You have $\lim_{n\to\infty} \sum_{k=1}^n \frac{1}{\sqrt{n^3 + 3k}}$. Hence you cannot move the limit inside the sum, as the sum depends on the limit. –  George V. Williams Jan 12 '13 at 3:48
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4 Answers

up vote 8 down vote accepted

Your evaluation is incorrect. For instance, consider the following. We have $$\underbrace{\dfrac1n + \dfrac1n + \dfrac1n + \cdots + \dfrac1n}_{n \text{ times}} = 1$$ If we were to apply an argument similar to yours, since $\lim_{n \to \infty} \dfrac1n =0$, we have $$1 = \lim_{n \to \infty} 1 = \lim_{n \to \infty} \left(\dfrac1n + \cdots + \dfrac1n \right) = \lim_{n \to \infty} \dfrac1n + \cdots +\lim_{n \to \infty} \dfrac1n = 0 + \cdots + 0 = 0$$which is clearly false.

The fact that "the limit of the sum is sum of the limits" is only true for finite sums and not for infinite sums i.e. $$\lim_{n \to \infty} \sum_{k=1}^{m} f_k(n) = \sum_{k=1}^m \lim_{n \to \infty} f_k(n)$$ is true only when $\lim_{n \to \infty} f_k(n)$ exists as a a real number and more importantly when '$m$' is a constant natural number independent of $n$.

Note that $$ \sum_{k=1}^n \dfrac1{\sqrt{n^3+3n}} \leq \sum_{k=1}^n \dfrac1{\sqrt{n^3+3k}} \leq \sum_{k=1}^n \dfrac1{\sqrt{n^3+3}}$$ Hence, we get that $$ \dfrac{n}{\sqrt{n^3+3n}} \leq \sum_{k=1}^n \dfrac1{\sqrt{n^3+3k}} \leq \dfrac{n}{\sqrt{n^3+3}} < \dfrac1{\sqrt{n}}$$ Now use squeeze theorem to obtain the limit.

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$n^2$ should be $n^3$. –  Primo Jan 12 '13 at 3:53
    
@Primo Thanks. corrected. –  user17762 Jan 12 '13 at 3:55
    
@Marvis Is strictly necesary $\frac{1}{\sqrt{n}}$? Or can be it ommited since both sides limits are 0? –  Alejandro Jan 12 '13 at 4:30
    
@Alejandro You can conclude the limit is $0$ even without $\dfrac{n}{\sqrt{n^3+3}} < \dfrac1{\sqrt{n}}$. I wrote $\dfrac{n}{\sqrt{n^3+3}} < \dfrac1{\sqrt{n}}$ since it is easy to show that $\dfrac1{\sqrt{n}}$ goes to $0$. –  user17762 Jan 12 '13 at 4:34
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The answer is correct, the reasoning is not. First we solve the problem.

The sum has $n$ terms. Each term is $\lt \frac{1}{n^{3/2}}$. It follows that if $S_n$ is our sum, then $$0\lt S_n \lt \frac{n}{n^{3/2}}=\frac{1}{n^{1/2}}.$$ Now let $n\to\infty$. Note that $\frac{1}{n^{1/2}}\to 0$, so by Squeezing $S_n\to 0$.

Remark: It is true that each term in the sum $S_n$ approaches $0$. However, the number of terms in the sum increases. Note for example that the sum $$T_n=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}$$ does not approach $0$, for clearly the sum is always $\ge \frac{1}{2}$. Yet if we applied the "$0+0+\cdots+0$" reasoning, we would conclude incorrectly that $\lim_{n\to\infty} T_n =0$.

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As $n$ grows, the number of terms increases. So you need extra care.

The sum is equal to $\sum_{k=1}^n \frac{1}{\sqrt{n^3+3k}}$.

We have $\left | \sum_{k=1}^n \frac{1}{\sqrt{n^3+3k}}\right| \leq n \cdot \frac{1}{\sqrt{n^3}}=\frac{1}{\sqrt{n}}$.

Now you can take a limit and the limit is $0$.

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Write the sum as

$$\sum_{k=1}^{n} \frac{1}{\sqrt{n^3 + 3 k}} $$

and note that

$$\frac{1}{\sqrt{n^3 + 3 k}} = \frac{1}{n^{3/2}} - \frac{3}{2} \frac{k}{n^{9/2}} + O \left ( \frac{k^2}{n^{15/2}} \right ) $$

When summing, the terms above vanish as $n^{-3/2}$, $n^{-5/2}$, etc. Draw your conclusion from that.

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