# When is the limit of a sum equal to the sum of limits?

I was trying to solve a problem and got stuck at the following step:

Suppose ${n \to \infty}$ .

$$\lim \limits_{n \to \infty} \frac{n^3}{n^3} = 1$$

Let us rewrite $n^3=n \cdot n^2$ as $n^2 + n^2 + n^2 + n^2 \dots +n^2$,$\space$ n times.

Now we have

$$\lim \limits_{n \to \infty} \frac{n^3}{n^3} = \frac {n^2 + n^2 + n^2 + n^2 + n^2 \dots +n^2}{n^3}$$

As far as I understand, we can always rewrite the limit of a sum as the sum of limits ...

$$\dots = \lim \limits_{n \to \infty} \left(\frac{n^2}{n^3} + \frac{n^2}{n^3} + \dots + \frac{n^2}{n^3}\right)$$

...but we can only let ${n \to \infty}$ and calculate the limit if all of the individual limits are of defined form (is this correct?). That would be the case here, so we have:

$= \dots \lim \limits_{n \to \infty} \left(\frac{1}{n} + \frac{1}{n} + \dots + \frac{1}{n}\right) =$[ letting ${n \to \infty}]$ $= 0 + 0 + \dots + 0 = 0$

and the results we get are not the same.

Where did I go wrong?

• Keep in mind that the number of summands in not fixed. Jan 3, 2016 at 19:20
• "We can always rewrite the limit of the sum as the sum of the limits." As you just found, that is not true. One can not always change nested limits around. For additional counterexample, consider $1 = \lim\limits_{x\to 0} \lim\limits_{y\to 0} x^y \neq \lim\limits_{y\to 0} \lim\limits_{x\to 0} x^y = 0$. You may be interested in reading about en.wikipedia.org/wiki/Fubini's_theorem Jan 3, 2016 at 19:22
• You cannot replace only a part of the expression with its limit, that's all. You have more and more terms, each getting smaller and smaller. Informally speaking, you do not have $n$ terms equal to $0$ but, so to say, an infinity of $0$s. Jan 3, 2016 at 19:23
• One way to look at it this way is to write $\lim_{n\to \infty}\frac{n^3}{n^3}=\lim_{n\to \infty}\frac1n \sum_{k=1}^n(1)$, which is the Riemann sum for $1$ over an interval of unit length. Of course, the value of the integral is $1$. Jan 3, 2016 at 19:25
• @JMoravitz This confuses me, because I thought that one can, in the case of $\lim \limits_{x \to \infty} f(x)$, *always* swap $lim$ and $f(x)$, under the condition that $f(x)$ is a continuous function. Thank you for the suggestion for further reading!
– 0lt
Jan 3, 2016 at 19:36

Because the number of terms goes up exactly as the size of each term goes down.

Specifically $$\lim \limits_{n \to \infty} \Big(\underbrace{\frac{1}{n} + \frac{1}{n} + \dots + \frac{1}{n}}_{n\text{ times}}\Big) = \lim \limits_{n \to \infty} \sum_{i=1}^n \frac 1n$$

Does that help?

• @Battani That proposed edit was incorrect. I realize it might look weird to not have any $i$'s in the summand but it really is supposed to be $\sum\limits_i \frac 1n = \frac 1n\sum\limits_i 1 = \frac 1n(n) = 1$.
– user137731
Jan 3, 2016 at 19:29
• If we pull out the $\frac{1}{n}$ expression in front of the sum, we are left $\lim \limits_{n\to \infty}\frac1n \sum_{k=1}^n1$ as mentioned in one of the comments above. The limit is 1, because I have an infinite sum of terms that are all slightly bigger than 0. Did I understand this correctly?
– 0lt
Jan 3, 2016 at 19:31
• Yep.$\ \ \ \ \$
– user137731
Jan 3, 2016 at 19:32
• Is there a general rule for rewriting the limit of a sum as the sum of limits?
– 0lt
Jan 3, 2016 at 19:33
• Assuming you have a finite sum, you can do so if each of the limits in the expression converges. It gets a little more complicated though with infinite sums.
– user137731
Jan 3, 2016 at 19:34

The problem you have described is common and there are three possible scenarios to cover:

1. Number of terms is finite and independent of $$n$$: The limit of a sum is equal to sum of limits of terms provided each term has a limit.
2. Number of terms is infinite: Some sort of uniform convergence of the infinite series is required and under suitable conditions the limit of a sum is equal to the sum of limits of terms.
3. Number of terms is dependent on $$n$$: This is the case which applies to the question at hand. The problem is difficult compared to previous two cases and a partial solution is provided by Monotone Convergence Theorem. Sadly the theorem does not apply to your specific example. But it famously applies to the binomial expansion of $$(1+n^{-1})^n$$ and gives the result $$e=1+1+\frac{1}{2!}+\frac{1}{3!}+\dots$$