# Summation with no argument

1. Feb 24, 2016

### BOAS

Hello,

in my QM class we arrived at the expression $\langle \hat{H} \rangle = \Sigma_{even n} |C_n|^2 E_n = \frac{24}{n^2 \pi^2} \frac{\hbar^2}{2m} \frac{n^2 \pi^2}{L^2}$.

The n terms cancel and we are left with $\langle \hat{H} \rangle = \frac{12 \hbar^2}{mL^2} \Sigma_{even n} 1$.

My lecturer said that this sum is infinity, since the number of even integers is infinity. Why is this the case when there are no n terms for the sum to act upon?

$\Sigma_{even n} n = \infty$, but I don't understand why $\Sigma_{even n} 1 = \infty$

2. Feb 24, 2016

### Samy_A

If you add 1 10 times, you get 10.
If you add 1 100 times, you get 100.
If you add 1 1000 times, you get 1000.
If you add 1 10000 times, you get 10000.

See why $\displaystyle \lim_{K\rightarrow +\infty} \sum_{n=0}^K 1$ is $+\infty$?

3. Feb 24, 2016

### BOAS

I understand what the sum to infinity means.

I don't think I have come across a sum over some index, where the index is not present and yet we still perform a sum.

$\Sigma_{n} n$ and $\Sigma_n 1$ don't appear to be the same thing, and yet they are treated the same. That's what I don't really get. I can accept that that is how things are done, but was wondering if there was a reason.

4. Feb 24, 2016

### Samy_A

While often the term being summed depends on the summation index n, there is no reason why this should always be the case.
$\displaystyle \sum_{n=0}^\infty 1=+\infty$ is a straightforward application of the definition of an infinite sum, or series.
The partial sums get as big as you want when you keep adding 1's, so the series goes to infinity.

5. Feb 24, 2016

### BOAS

Ok, if i'm meant to read $\displaystyle \sum_{n=0}^\infty 1$ as the sum of 1 from zero to infinity, that's fine.

6. Feb 24, 2016

### Edgardo

For a sequence $(a_n)_{n \in \mathbb{N}}$ you can form the series $\sum_{n=0}^\infty a_n = a_0 + a_1 + a_2 + \dots$.

If you take the sequence $a_n = 1$ and plug it into the expression above,
then you get the series $\sum_{n=0}^\infty a_n = \sum_{n=0}^\infty 1 = 1 + 1 + 1 \dots$.

Or consider the partial sum $s_k = \sum_{n=0}^k a_n = a_0 + a_1 + \dots + a_k$. If you plug in the sequence $a_n = 1$, then you get the partial sum $s_k = \sum_{n=0}^k a_n = a_0 + a_1 + \dots + a_k = \underbrace{1 + 1 + \dots + 1}_{\text{k+1 summands}}$.

Last edited: Feb 24, 2016
7. Feb 24, 2016

### PeroK

In fact:

$\displaystyle \sum_{n=1}^m a$

is multiplication of $a$ by $m$

8. Feb 24, 2016

### DrewD

It might be useful to think about the definition of the summation function as $\sum_{n=0}^Nf(n)=f(0)+f(1)+f(2)+\ldots+f(N)$. Then $f(n)=1$ for your example and the definition works. Usually this is written with $f(n)=a_n$, but for some reason when I saw it with $f(n)$ for the first time it clicked better for me.