# Evaluate limit of the series: $\lim_{n\to\infty} \left[\frac{1}{n^{2}} + \frac{2}{n^{2}} + \frac{3}{n^{2}} + \cdots + \frac{n}{n^{2}}\right]$

I am pretty confident that the following limit is $$0.5$$:

$$\lim_{n\to\infty} \left[\frac{1}{n^{2}} + \frac{2}{n^{2}} + \frac{3}{n^{2}} + \cdots + \frac{n}{n^{2}}\right]=\lim_{n\to\infty} \left[\frac{1+2+3+ \cdots +n}{n^{2}}\right]=\lim_{n\to\infty} \left[\frac{n^2+n}{2n^{2}}\right]=\frac{1}{2}$$

However one of the students argued that if we write limit of sum as sum of individual limits, it will be zero. Why we cannot write limit of sum as sum of limits in this case?

I've been taught that if individual limits exist, the limit of sum is equal to the sum of limits. It would be helpful to get an explanation or a reference to similar rules for limits of series.

• that's because $n$ tends to infinity, so the limit of the sum is not equal to the sum of the limits – Alex Mar 27 at 12:58
• Because if we could write that every limit of sum is sum of limit, thousands of professors will be out of work. – uniquesolution Mar 27 at 13:06
• Just as an aside, this is a simple example of why we need an increasing sequence of functions in Beppo Levi's theorem. Write $$f_n(k)=\begin{cases} {k\over n^2},&1\leq k\leq n\\0,&k>n\end{cases}$$ and the $f_n$ decrease monotonically to $0$. – saulspatz Mar 27 at 13:24
• As $n$ tends to infinity, the individual limits will become $0$, but there will be infinitely many of them. So that’s a case of $0\cdot\infty$. – Jonas De Schouwer Mar 27 at 14:00

However one of the students argued that if we write limit of sum as sum of individual limits, it will be zero. Why we cannot write limit of sum as sum of limits in this case?

For the sum of two sequencesn we have the property "limit of the sum is the sum of limits" (if both sequences have a limit) and by repeatedly applying this, we have this property for any finite number of sequences (terms).

As often is the case, you cannot simply extend this to the infinite case; i.e. you cannot assume the same property will hold when the number of sequences (terms) is not finite.

A simpler counterexample would be the sum of $$n$$ terms, all equal to $$\tfrac{1}{n}$$; obviously we have: $$\underbrace{\frac{1}{n}+\frac{1}{n}+\ldots+\frac{1}{n}}_{\mbox{n terms}} = \frac{n}{n}=1$$ but every individual sequence (term) clearly tends to $$0$$: $$\frac{1}{n} \stackrel{n\to \infty}{\longrightarrow} 0$$.

You can prove by induction on $$k$$ that, if $$k$$ functions of $$n$$ each converge as $$n\to\infty$$, the sum of their limits is the limit of their sum. The expression you're dealing with isn't an example of that. Instead there are infinitely many functions, of which an $$n$$-dependent number are used in the sum. There's no way to bridge this gap, by induction or otherwise. Indeed, the example we're discussing proves that's impossible, as the limit is indeed $$\frac12$$.

Just another way to "convince" you that the limit is $$\frac{1}{2}$$:

The sums are Riemann sums:

• $$\left[\frac{1}{n^{2}} + \frac{2}{n^{2}} + \frac{3}{n^{2}} + \cdots + \frac{n}{n^{2}}\right] = \sum_{k=1}^n\left( \frac{k}{n} \cdot \frac{1}{n}\right) \stackrel{n\to \infty}{\longrightarrow} \int_0^1 x \;dx = \frac{1}{2}$$

Riemann sums serve as a perfect example that interchanging infinite summation and limits of members of the involved sums cannot work in general.

Possibly interesting further readings could be something about interchanging limit operations or more specifically about interchanging summation and limits.