Is there a way to prove continuity without using epsilon-delta? I'd like to know if there is an easier way of provining continuity instead of using the epsilon-delta criteria? I cannot understand it because it's way too complicated for me...
There is no workaround? Like converting the series to a function, then prove convergence on the function. If the converted function is continuous, the series will be continuous as well. Something like that would be possible?
Or there are other ways which are a bit easier?
Sorry for asking a question like that but I couldn't find anything on the internet. I really hope there is another way...
 A: This answer is about the continuity of functions, since the comments under the question indicate that the question was really about continuity of functions rather than convergence of series.
The continuity of a function $f:A\to\mathbb R$, 
where $A \subseteq \mathbb R$,
is usually defined in terms of limits of the form $\lim_{x\to a} f(x)$.
Limits of functions are usually defined using quantifiers over
variables $\epsilon$ and $\delta$.
The reason the definition is important is that it applies to every function
you could ever look at.
It is possible to write the criteria for continuity differently
so that they still apply to every function, either by proof or by definition.
This other question gives an example of an
alternative way to define continuity.
But the application of this definition to an actual function is
(I think) at least as complicated as the $\epsilon$-$\delta$ method.
I think if you want a method you can apply to any function, you need to have
enough mathematical understanding to do $\epsilon$-$\delta$ proofs reliably.
The alternatives will require just as much sophistication.
For many functions, however, you can avoid having to write out
the $\epsilon$-$\delta$ method if you are allowed to apply a few known facts.
For example, if $f$ and $g$ are both continuous functions over the
domain $A$, then $f + g: x \mapsto f(x) + g(x)$ is also a continuous
function over the domain $A$.
There are several useful facts like this by which you can combine
known continuous functions to show that other functions are continuous;
page 2 of this document, for example, shows some of them.
Realize that someone had to apply $\epsilon$-$\delta$ proofs or similarly
difficult methods in order to prove all of those facts in the first place.
But once you know they are true, they make it very easy to show that
(for example) the function $x \mapsto \frac{x^2}{x+1}$ on the domain $[0,\infty)$ is continuous.
A: Based on @Dave L. Renfro's comment, I am inclined to include a detailed example, since that might help more than or at least supplement my explanation. This one is moderately tricky, if I remember and execute this correctly (it's been awhile).
Say $f(x)=x^{2}$ and we want to find a general $\delta$ that holds for all $x \in \mathbb{R}$ such that for alll $\epsilon>0$ we have that $\lvert x-a \rvert<\delta$ implies that $\lvert x^{2}-a^{2} \rvert<\epsilon$. 
Now, to start, note that we can factor our inequality 
$$
\lvert x^{2}-a^{2} \rvert=\lvert x-a\rvert\lvert x+a\rvert 
$$
now notice that we have
$$
\lvert x-a \rvert<\delta \text{ and } \lvert x-a\rvert\lvert x+a\rvert<\epsilon
$$
so we can write 
$$
 \lvert x-a\rvert<\frac{\epsilon}{\lvert x+a\rvert}$$
and furthermore, say that we were to assume $\delta$ satisfies
$$
\lvert x-a\rvert<\delta \leq \frac{\epsilon}{\lvert x+a\rvert}
$$
and 
$$
\delta \leq \frac{\epsilon}{\lvert x+a\rvert} \implies  \lvert x-a\rvert\lvert x+a\rvert<\epsilon
$$
but notice that $\delta$ depends on $x$, and ideally, we want $\delta$ to be a function of $a$ and $\epsilon$ only. To sidestep this issue, we assume that $\delta <1$ then by choosing a $\delta \leq 1$ we have that $\lvert x-a \rvert < 1$ whenever $\lvert x-a\rvert<\delta$.
Now note that $\lvert x-a \rvert \leq 1$ is equivalent to saying that  $x$ is within a distance of $1$ of $a$ which implies that for $\lvert x+a\rvert$ we have 
$$
\lvert x+a \rvert +\lvert x-a\rvert<1+\lvert x+a\rvert  \implies \lvert x+a \rvert <\lvert 2a \rvert +1
$$
so now we have a bound on $\lvert x+a \rvert$ which only depends on $a$ so we can write
$$
\lvert x-a\rvert < \delta=\min\left\{1,\frac{\epsilon}{\lvert 2a \rvert+1} \right\}<\frac{\epsilon}{\lvert x+a\rvert}
$$
Which shows that $f(x)=x^{2}$ is continuous for any $a \in \mathbb{R}$.
A: You bet there is! To prove the continuity of $y=f(x)=\frac{x^2}{x+1}$ on positive reals, just show that every infinitesimal $x$-increment always leads to an infinitesimal change in $y$, which is the Cauchy-Robinson definition of continuity.  Namely, if $\alpha>0$ is infinitesimal then 
$$f(x+\alpha)-f(x)= \frac{(x+\alpha)^2}{x+\alpha+1}-\frac{x^2}{x+1}=\frac{(x+\alpha)^2(x+1)-(x+1+\alpha)x^2}{ (x+1+\alpha)(x+1) }.
$$
Multiplying out, we get
$$
f(x+\alpha)-f(x)= \frac{x^3+2x^2\alpha+\alpha^2x + x^2+2x\alpha+\alpha^2-x^3-x^2-\alpha x^2}{ (x+1+\alpha)(x+1)}.
$$
Simplifying, we get
$$
f(x+\alpha)-f(x)= \frac{2x^2\alpha+\alpha^2x +2x\alpha+\alpha^2-\alpha x^2}{(x+1+\alpha)(x+1)}=\alpha\frac{2x^2+\alpha x +2x+\alpha- x^2}{(x+1+\alpha)(x+1)}.
$$
This is a product of an infinitesimal $\alpha$ by a finite number $\frac{2x^2+\alpha x +2x+\alpha- x^2}{(x+1+\alpha)(x+1)}$ which is therefore infinitesimal. QED. For details on the rules governing manipulation with infinitesimals see Elementary calculus.
A: Topologically you can argue that given two topological spaces $(A, \tau_{A}),(B, \tau_{B})$ a function is continuous if for  $f:A \to B$ we have $U \in \tau_{B} \implies f^{-1}(U) \in \tau_{A}$. 
Edit: 
First of all, given an arbitrary space $X$, a metric is a function $d:X \times X \to [0,\infty)$ where together the space and the metric $(X,d)$ form a metric space (and also note that a metric space is a topological space since metrics induce topologies). To be a metric, $d$ must satisfy some intuitive properties namely that 
$$d(x_{1},x_{2})=0 \implies x_{1}=x_{2}$$
and
$$
d(x_{1},x_{2})=d(x_{2},x_{1})
$$
and finally the triangle inequality
$$
d(x_{1},x_{3}) \leq d(x_{1},x_{2})+d(x_{2},x_{3})
$$
Now we usually denote the metric on $\mathbb{R}$ by $\lvert \cdot \rvert$ which for $x_{1},x_{2} \in \mathbb{R}$ we have just $\lvert x_{1} \rvert =d(0,x_{1})$ and $\lvert x_{1}-x_{2} \rvert=d(x_{1},x_{2})$. 
Now for $\epsilon-\delta$ proofs, we have the general statement in plain English that for a function $f:\mathbb{R} \to \mathbb{R}$ we have that
$$
\lim_{x \to x_{0}}f(x)=L
$$
if for $\epsilon >0$ there exists a $\delta > 0$ such that $\lvert x- x_{0}\rvert <\delta$ imples that $\lvert f(x)-L \rvert<\epsilon$
Now what is really going on here? Well first note that we can choose our value of epsilon since it is arbitrary and if $f$ is continuous, then there will be a corresponding value of $\delta$ that guarantees that $\lvert x-x_{0} \rvert<\delta$. This means that $\delta$ IS A FUNCTION OF EPSILON, which is the part that so many teachers do not explicitly mention, thus leading to so much confusion. Again, just to make that as explicit as possible, $\delta$ is uniquely determined by our choice of $\epsilon$. Now what exactly is $\epsilon$? Well, recall from algebra that
$$
\lvert f(x)-L \rvert<\epsilon \text{ is equivalent to writing } L-\epsilon<f(x)<L+\epsilon
$$
so $\epsilon$ quantifies the degree to which we "squeeze" the interval $(L-\epsilon,L+\epsilon)$ and $\delta$ quantifies the degree to which we squeeze the interval $(x_{0}-\delta,x_{0}+\delta)$ based on our choice of $\epsilon$. Now, this might seem backwards since $x$ is the independent variable and the value of $f(x)$ depends on our choice of $x$ but that is actually exactly why we need this framework. Let's say we reversed the statement and we squeezed $(x_{0}-\delta,x_{0}+\delta)$ by choosing $\delta$ instead of epsilon, then the fact that there is some value $\epsilon$ such that $\lvert f(x)-L\rvert <\epsilon$ is trivial for any (bounded) function. Basically, the intuition for continuity is that continuity implies that when two values $x_{1}$ and $x_{2}$ are "close" (an inherently topological notion) to each other then $f(x_{1})$ and $f(x_{2})$ have to be close to each other as well. Thus, by squeezing the interval $(L-\epsilon, L+\epsilon)$ then for a continuous function, we know that the interval $(x_{0}-\delta,x_{0}+\delta)$ will be squeezed to a distance of $\delta$ as a function of  $\epsilon$. 
Now see if you can "reverse engineer" the $\epsilon-\delta$ definition of continuity to arrive at the topological definition I provided above.
A: You could try reasoning within a framework which extends the reals with actual infinitesimals (I am aware of three such frameworks).
A: Usually, there are simple ways to show continuity.
Composition of continuous functions is continuous
If you build your function out of continuous functions, it is automatically continuous. For example, consider the function:
$$ f(x) = \sqrt{1 + x^2} $$
Constants, addition, squaring, and square roots are all continuous on their domain, and this implies $f$ is continuous.
Compute the limit
The usual definition of continuity is not $\epsilon-\delta$: it is in terms of limits. You spend a lot of time in your calculus classes learning how to compute them.
For example, consider the function
$$ \mathrm{sinc}(x) = \begin{cases} \frac{\sin x}{x} & x \neq 0 \\ 1 & x = 0 \end{cases} $$
By the method above, we can immediately see that $\mathrm{sinc}$ is continuous whenever $x \neq 0$. To show continuity at $x=0$, we just compute the limit.

Goal: prove $\mathrm{sinc}(0) = \lim_{x \to 0}\mathrm{sinc}(x) $

The left hand side is $1$. The right hand side can be computed by
$$ \lim_{x \to 0} \mathrm{sinc}(x) = \lim_{x \to 0} \frac{\sin x}{x} = 1 $$
Both sides are equal, and so $\mathrm{sinc}$ is continuous at zero as well, so it's continuous everywhere.
