Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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.

share|cite|improve this answer
$n^2$ 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

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 $\lim_{n\to\infty} T_n =0$.

share|cite|improve this answer

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 $0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.