Estimating infinite series on Desmos How would I go about estimating this infinite series on demos? $\sum^{\infty}_{n=1}\frac{10+\cos(n)}{n^2}$
I realize that graphing calculators such as Desmos are have an inability to graph infinity, so how would I deconstruct this series to allow a calculator to estimate it?
 A: Unfortunately you can't really get rid of the $\infty$ thing for desmos.
I would suggest creating a slider for some number $N\geq0$, and setting the step of the slider to $1$. Then define
$$A=\sum_{n=1}^{N}\frac{10+\cos(n)}{n^2}$$
The bigger and bigger you make $N$, the better the approximation. I usually like to do $N=100$ because it doesn't take too long with desmos. However you will get a more accurate result with $N=1000$. 
You could also replace $N$ with $x$ and watch it converge.

In case you are interested, here's how to find the exact value of part of this sum.
$$S=\sum_{n=1}^{\infty}\frac{10+\cos(n)}{n^2}=10\sum_{m=1}^{\infty}\frac1{m^2}+\sum_{n=1}^{\infty}\frac{\cos(n)}{n^2}$$
I know how to deal with the first one but not the second one.
We start with 
$$S_1=\sum_{n=1}^{\infty}\frac1{n^2}$$
For this we recall the definition of Fourier Series. For some $L>0$ and some function $f(x)$ which is continuous on $[-L,L]$, we have that for $-L\leq x\leq L$,
$$f(x)=\frac{a_0}2+\sum_{n=1}^{\infty}\bigg[a_n\cos\bigg(\frac{n\pi x}{L}\bigg)+b_n\sin\bigg(\frac{n\pi x}{L}\bigg)\bigg]$$
Where 
$$a_n=\frac1L\int_{-L}^{L}f(x)\cos\bigg(\frac{n\pi x}{L}\bigg)\mathrm{d}x$$
And 
$$b_n=\frac1L\int_{-L}^{L}f(x)\sin\bigg(\frac{n\pi x}{L}\bigg)\mathrm{d}x$$
Choosing $f(x)=x^2$ and $L=\pi$, we have (for $|x|\leq\pi$), 
$$x^2=\frac{a_0}2+\sum_{n=1}^{\infty}\bigg[a_n\cos(nx)+b_n\sin(nx)\bigg]$$
So we start with $b_n$:
$$b_n=\frac1\pi\int_{-\pi}^{\pi}x^2\sin(nx)\mathrm{d}x$$
With the substitution $x=-u$, we see that 
$$b_n=0$$
Great! Let's move on:
$$a_0=\frac1\pi\int_{-\pi}^{\pi}x^2\cos(0x)\mathrm{d}x=\frac1\pi\int_{-\pi}^{\pi}x^2\mathrm{d}x=\frac{2\pi^2}3$$
And 
$$a_n=\frac1\pi\int_{-\pi}^{\pi}x^2\cos(nx)\mathrm{d}x$$
I leave it up to you as a teeny little challenge to show that 
$$a_n=(-1)^n\frac{4}{n^2}$$
(Hint: $\cos(n\pi)=(-1)^n$ for integer $n$)
So we update our Fourier series:
$$x^2=\frac{\pi^2}{3}+4\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\cos(nx)$$
Again using $\cos(n\pi)=(-1)^n$ for integer $n$, we may plug in $x=\pi$ to see that
$$\pi^2=\frac{\pi^2}{3}+4\sum_{n=1}^{\infty}\frac{1}{n^2}$$
So we see that 
$$S_1=\frac{\pi^2}6$$
So we have that 
$$S=\frac{5\pi^2}{3}+\sum_{n=1}^{\infty}\frac{\cos(n)}{n^2}$$
you can use the $N$ slider trick I told you about for this remaining sum.
