calculuspower series
$$\int_{0}^{\infty} \mathrm{e}^{-x^2}\, dx = \frac{\sqrt \pi}{2}$$
$$\int_{0}^{\infty} \sum_\limits{k=0}^\infty\frac{(-1)^k(x^{2k})}{k!}dx=\frac{\sqrt{\pi}}2$$
I don't know much about this I just thought hey right we have the function what if we wrote it as the power series and integrated that and got the sum would it work? On a limb I saw this post
Are these the notes you took in class? ie.. do you just want some explanation on some of the details? Well for your first example, in the last step the $$1\over 5$$ is multiplied into the summation giving the term $$x^{n+1}\over {5^{n+1}(n+1)}$$. Then the power series is shifted by replacing n with n-1 (which explains the term you have written in the last step). However, in doing this, the series now starts at $n=1$ not $n=0$. Was that a typo? Clearly the term to the right of the last summation is undefined at $n=0$
For the second example (and I guess the first one too) the term $C$ is an arbitrary constant of integration. So to evaluate the definite integral in your second question, you can choose $C$ to be zero as any antiderivative can be used in the fundamental theorem of calculus.
In general, power series are nice because polynomials are by far the easiest functions to work with, and a power series allows you approximate a much harder function as a polynomial. There are only a few well known power series in elementary calculus (namely, $sin(x)$, $cos(x)$, $e^x$, and $1\over {x-1}$ ) but by differentiating and integrating you can produce power series for many, many functions. That's the whole point of this.
$$e^z=\sum_{n=0}^\infty \frac{z^n}{n!}\qquad\forall z\in\mathbb{C}$$
and so:
$$e^{-x^2}=\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{n!}$$
the problem is that the terms alternate between positive and negative, so if you cut the series off at any arbitrary point (which is going to be small compared to the infinite number of terms). To demonstrate this point look at the graph below:
The red line represents the actual $\exp(-x^2)$ function whilst blue is $n\in[0,5]$ and green is $n\in[0,10]$ and for your integral you want to integrate this series and then take the limit for $n\to\infty$ whilst will just accentuate any "small" error you see here
Best Answer
If you exchange the integral and the series blindly, the integral of each term is infinite, showing that it was not allowable to exchange the series and the integral.
Remember that $\int_0^\infty e^{-x^2}\,dx$ means $$ \lim_{N\to\infty}\int_0^N e^{-x^2}\,dx. $$ Now you can do (as you saw in the post you mention, because the convergence of the series is uniform on closed intervals) $$ \lim_{N\to\infty}\int_0^N\sum_{k=0}^\infty \frac{(-1)^k x^{2k}}{k!}\,dx =\lim_{N\to\infty}\sum_{k=0}^\infty \int_0^N\frac{(-1)^k x^{2k}}{k!}\,dx =\lim_{N\to\infty}\sum_{k=0}^\infty \frac{(-1)^k N^{2k+1}}{k!(2k+1)}. $$ And now you need to calculate the new series. The best method is to notice that it is the anti-derivative of $e^{-x^2}$ at $x=N$, so $$ \sum_{k=0}^\infty \frac{(-1)^k N^{2k+1}}{k!(2k+1)}=\int_0^Ne^{-x^2}\,dx. $$ Not super useful.