# Difference between $\lim_\limits{n\rightarrow\infty}\sum_\limits{k=1}^{n}a_k$ and $\sum_\limits{k=1}^{\infty}a_k$

Is there any difference between $\lim_{n\rightarrow\infty}\sum_\limits{k=1}^{n}a_k$ and $\sum_\limits{k=1}^{\infty}a_k$?

My example and thought:

Let $a_n=n$ where $n\in\mathbb{P}$. $\mathbb{P}$ is the set of all positive integer.

Then $$\lim_{n\rightarrow\infty}\sum_{k=1}^{n}a_k=\lim_{n\rightarrow\infty}\sum_{k=1}^{n}k=\lim_{n\rightarrow\infty}\frac{n(n+1)}{2}\rightarrow\infty$$

And

$$\sum_{k=1}^{\infty}a_k=\sum_{k=1}^{\infty}k=-\frac{1}{12}.$$

Am I thinking right?

• How do you define $\sum \limits_{k=1}^{\infty}a_k$? – Git Gud Mar 23 '14 at 14:34
• @GitGud Sum of all members of ${a_k}$. – Novice Mar 23 '14 at 14:35
• That's an infinite sum, it isn't defined. – Git Gud Mar 23 '14 at 14:36
• @GitGud Then, is $\sum \limits_{k=1}^{\infty}a_k$ wrong notation? Or we define $\sum \limits_{k=1}^{\infty}a_k:=\lim \limits_{n\rightarrow\infty}\sum \limits_{k=1}^{n}a_k$? – Novice Mar 23 '14 at 14:46
• @Novice, exactly right. Check your books. – vonbrand Mar 23 '14 at 15:07

What you are using in your example is a Zeta regularization of a divergent series. In that sense, even the equality sign is not really justified. $-1/12$ is not the sum in the traditional sense, it's a generalization and you have to specify what you are doing.
• I'm getting more confusing now. Git Gud said in the comment that infinite sum is not defined, but you are saying that infinite sums are defined as limits of their partial sums, which is $\sum \limits_{k=1}^{\infty}a_k:=\lim \limits_{n\rightarrow\infty}\sum \limits_{k=1}^{n}a_k$. I don't know what zeta regulation is though. I've just read the proof of sum of all natural number is $-1/12$ which has clearly proved. – Novice Mar 23 '14 at 15:04
• Ok, I've just read the definition of series, no reason to be a not well versed in mathematics at least in this case. If the sequence of partial sums of $\{a_n\}$; $\{S_n\}$ converges, then we define the infinite series $\sum \limits_{n=1}^{\infty}a_n:=\lim \limits_{n\rightarrow\infty}S_n$. And the sum of all natural number is also derived in traditional way. I've read the proof of it but it was clearly proved without using any contemporary mathematical theories. – Novice Mar 23 '14 at 15:35
• Let $S_1:=1-1+1-1+1-\cdots$. We can easily observe that $S_1=1-S_1$, $S_1=1/2$. Now, let $S_2:=1-2+3-4+5-6+7-\cdots$. Then $2S_2=S_1=1/2$, $S_2=1/4$. And let $S:=1+2+3+4+5+6+7+\cdots$. Consider $S-S_2=4S$, then $3S=-S_2=-1/4$. Therefore, $S=-1/12$. – Novice Mar 23 '14 at 15:44
• In this case, the "f the sequence of partial sums of {an}; {Sn} converges" does not apply. The series is divergent. If you want to play around with this idea, you can also look at the $1+q+q^2+q^3\cdots=1/(1-q)$ and use $|q|>1$. You are using a series outside its convergence radius (just like with $1+2+3+\cdots$), so in the traditional sense, the sum doesn't exist. You are extrapolating to get a generalized notion of the sum. – orion Mar 23 '14 at 16:13
Your example is a red herring; the standard way of interpreting the sum $\sum_{k=1}^\infty k$ is that it diverges to infinity, although there are some theories of divergent series which would assign the sum a different value. In any case, the two limits in your question are exactly equal: an infinite series is defined to be the limit of its partial sums.