$\sum_{i=1}^{n}{\frac{1}{(2n+i)^2}}=?$
How do I evaluate this? I am a beginner, so please be gentle with your professional answers...
By the way, I am asking this because I am trying to darboux-integrate something and got this times $n$ as lower darboux sum. So I basically need the limit of this as $n\to\infty$.
You can make the substitution $j = 2n+i$:
$$ \sum_{i=1}^{n}{\frac{1}{(2n+i)^2}} = \sum_{j=2n+1}^{3n} \frac{1}{j^2} = \sum_{j=1}^{3n} \frac{1}{j^2} - \sum_{j=1}^{2n} \frac{1}{j^2}. $$
If you want to evaluate for specific $n$, $$\displaystyle \sum_{j=1}^N \frac{1}{j^2} = \left.\frac{d^2}{dx^2}\left(\ln(\Gamma(x))\right)\right|_{x=N+1}$$ where $\Gamma(x)$ is the gamma function; these are the generalized Harmonic numbers in power $2$.
The limit is easier, since $\displaystyle \sum_{j=1}^{\infty} \frac{1}{j^2}$ is a convergent series. Indeed,
$$0 \leq \lim_{n\rightarrow \infty} \sum_{j=2n+1}^{3n} \frac{1}{j^2} \leq \lim_{n\rightarrow \infty} \sum_{j=2n+1}^{\infty} \frac{1}{j^2} = 0.$$