# Infinity in infinite series

We define infinite series as summation of terms in sequence. where sequence is defined to function from natural number to real numbers $$f:\mathbb{N} \rightarrow \mathbb{R}$$. But $$\infty$$ does not belong to natural numbers right? so what does $$\infty$$ as upper limit represent in this $$\sum_{n=1}^{\infty} a_n$$

I am confused because consider Cauchy Criterion which states as follows

A series $$\sum_{n=1}^{\infty} a_n$$ converges iff $$\forall \epsilon>0$$ $$\,$$ $$\exists N \in \mathbb{N}$$ $$\,$$ $$\forall n>m\geq N$$ $$\,$$ $$|\sum_{k=m+1}^{n} a_k| < \epsilon$$

can I take upper limit to $$n = \infty$$ that is can i say $$|\sum_{k=m+1}^{\infty} a_k| < \epsilon$$ ?

I think answer is no because $$\infty$$ does not belong to natural numbers.But we seem to be doing that when we write infinite series notation

• $1\leq n<\infty$, like in the range of Python. – logarithm May 22 '19 at 11:22
• In the Cauchy criterion the $\forall n> m\geq N$ doesn't include $n=\infty$. It is not needed. However, if you write the same proposition including those sums, the resulting proposition is equivalent (on the real numbers). This is because if the Cauchy condition is satisfied, then the object called $\sum_{m+1}^{\infty}a_n$ exists and its absolute value is $\leq \epsilon$. Taking a smaller $\epsilon$ you can ensure the strict inequality. – logarithm May 22 '19 at 11:24

The $$\infty$$ symbol in the summation is just notation. It simply means, '$$n$$ may be arbitrary large', not that $$n$$ ever takes on an 'infinite' value—whatever that would mean.

It is for this reason that you can't start substituting $$\infty$$ into definitions like the Cauchy criterion for convergence. None of our definitions apply to it, because it isn't a number.

As an example what is the value of $$\frac{1}{\infty}$$ ? Well, we know that the sequence $$(\frac{1}{n})$$ approaches $$0$$, so maybe we can say that $$\frac{1}{\infty} = 0$$. Now consider that we also have $$\frac{x}{x} = 1$$ for all nonzero $$x \in \mathbb{R}$$.

What is $$\frac{\infty}{\infty}$$, then? From our first fact we would seem to say $$0$$, but from our second fact it seems clear that it must be $$\infty$$. We have a paradox. And no wonder, because what we are manipulating is not even defined.

As you say, 'infinity' is not a natural number, and in fact nor is it a real number. It is simply the notion of unbounded growth, and perhaps it is confusing that we use its symbol in place of a number of the kind we are used to.

We define infinite series as summation of terms in sequence.

It is not exactly clear what you mean here, but we can define the value of an infinite series as the value that the sequence of its partial sums approaches. Since we already have a very rigorous and well-defined notion of convergence for sequences, this is now a rigorous definition of convergence of a series.

The $$\infty$$ symbol in your summation notation just means that we are finding the value that the sequence of partial sums converges to, rather than just the partial sum after a certain finite number of terms.

You are correct in your statements regarding the infinity not being a number.

So, what is it ?

We define a sequence approaching infinity if the terms grow without bound.

Same goes with series, when we add up terms of an infinite sequence we are trying to find the limit of the sequence of partial sums as the index grows without bound.

Therefore infinity is not a number but we need it to express the growth without bound.

In the notation, $$\sum_{n=0} ^\infty a_n$$, the $$\infty$$ is there just to say the is no upper limit on the index, n. It is not intended to say that n ever takes on value "$$\infty$$".

A series $$\sum_{n=1}^{\infty} a_{n}$$ converges iff $$\forall \epsilon>0 \exists N \in \mathbb{N} \forall n>m \geq N\left|\sum_{k=m+1}^{n} a_{k}\right|<\epsilon$$

Is the same as: A series $$\sum_{n \in \mathbb{N}} a_{n}$$ converges iff $$\forall \epsilon>0 \exists N \in \mathbb{N} \forall n,m \in \mathbb{N}: n>m \geq N \implies \left|\sum_{k=m+1}^{n} a_{k}\right|<\epsilon$$