Does sequence ${ s }_{ n }=\sum _{ k=0 }^{ n }{ \frac { 1 }{ { 1+k }^{ 2 } } } $ converge I'm reviewing some Calculus 1 convergence stuff. I want to decide wether ${ s }_{ n }:=\sum _{ k=0 }^{ n }{ \frac { 1 }{ { 1+k }^{ 2 } }  } $
is convergent or not.
Since the sequence is monotnically increasing and has an upper and lower bound it is convergent, but im failing to make the right estimates
 A: Note that
$$
0<\sum _{ k=1 }^{ \infty }\frac { 1 }{ { 1+k }^{ 2 } }\leq \sum _{ k=1 }^{ \infty }\frac {1}{k^2}.
$$
You should be able to tell whether the last series has a finite value. 
A: Hint: Consider the comparison test of series. 
A: Note $k^2+1>k^2$. So $\frac{1}{k^2+1}<\frac{1 }{k^2}$. And now use comparison test. 
A: You can do much more than proving it is convergent (trivial by comparison with $1+\sum_{k\geq 1}\frac{1}{k^2}=1+\frac{\pi^2}{6}$), you can compute $\lim_{n\to +\infty}s_n$ in a explicit way. Since
$$\forall k\in\mathbb{N}^+,\qquad \frac{1}{k^2+1} = \int_{0}^{+\infty}\frac{\sin(k x)}{k}\,e^{-x}\,dx \tag{A}$$
follows by integration by parts, and $\sum_{k\geq 1}\frac{\sin(k x)}{k}$ is the Fourier series of the sawtooth wave, which is $2\pi$-periodic and equals $\frac{\pi-x}{2}$ over the interval $x\in(0,2\pi)$, we have
$$ \sum_{k\geq 1}\frac{1}{k^2+1}=\int_{0}^{2\pi}\frac{\pi-x}{2}\left[e^{-x}+e^{-(x+2\pi)}+\ldots\right]\,dx=\frac{1}{1-e^{-2\pi}}\int_{0}^{2\pi}\frac{\pi-x}{2}\,e^{-x}\,dx\tag{B} $$
and by integration by parts
$$ \sum_{k\geq 1}\frac{1}{k^2+1}=\frac{1}{1-e^{-2\pi}}\cdot\frac{(\pi+1)e^{-2\pi}+(\pi-1)}{2}=\color{red}{\frac{\pi\coth\pi-1}{2}}.\tag{C}$$
$(C)$ is also a consequence of the Poisson summation formula, since the Cauchy distribution and the Laplace distribution are conjugated via the Fourier transform.
A: 1) Sequence is monotonically increasing .
2)Need to find an upper bound:
$\sum_{k=1}^{n} \dfrac{1}{1+k^2}\le$
$\sum_{k=1}^{n}\dfrac{1}{k^2}$.
Consider:
$\sum_{k=2}^{n}\dfrac{1}{k^2} \le$
$\sum_{k=1}^{n-1}\dfrac{1}{k(k+1)} =$
$\sum_{k=1}^{n-1}[\dfrac{1}{k} - \dfrac{1}{k+1}]=$
$1 - \dfrac{1}{n}.$
Bounded above.
Sum converges.
