Sums $\sum_{n=1}^{N}\sqrt{4n+1}$

Question

I need to find sum of the first N terms of the sequence whose nth term is as follow :

T(n)= $\sqrt{4*n+1}$

So the sequence is : $\sqrt{5}$,$\sqrt{9}$,$\sqrt{13}$,$\sqrt{17}$,$\sqrt{21}$......

Please help how to find it for a given N.

Answer 1

There is an answer which is not simple at all since $$\sum_{i=1}^n \sqrt{4i+1}=2 \left(\zeta \left(-\frac{1}{2},\frac{5}{4}\right)-\zeta \left(-\frac{1}{2},n+\frac{5}{4}\right)\right)$$ in which appears Hurwitz $\zeta$ function.

However, you could notice that $$2\sqrt{i}<\sqrt{4i+1}<2\sqrt{i+1}$$ and so $$2H_{n}^{\left(-\frac{1}{2}\right)}<\sum_{i=1}^n \sqrt{4i+1}<2H_{n+1}^{\left(-\frac{1}{2}\right)}$$ where appear the generalized harmonic numbers.

For large values of $m$, an approximation is $$H_{m}^{\left(-\frac{1}{2}\right)}=\frac{2 m^{3/2}}{3}+\frac{\sqrt{m}}{2}+\zeta \left(-\frac{1}{2}\right)+O\left(\left(\frac{1}{m}\right)^{3/2}\right)$$

For $n=10$, the exact summation leads to $46.1629$ while the expression on the left is $44.9102$ and the one on the right is $51.5447$

Added later

The approximation can be improved writing $$\sqrt{4i+1}=2 \sqrt{i}+\frac{\sqrt{\frac{1}{i}}}{4}-\frac{1}{64} \left(\frac{1}{i}\right)^{3/2}+O\left(\left(\frac{1}{i}\right)^{5/2}\right)$$ and expanding again the harmonic numbers. As a result comes the approximation $$\sum_{i=1}^n \sqrt{4i+1}\approx \frac{4 n^{3/2}}{3}+\frac{3 \sqrt{n}}{2}+\left(2 \zeta \left(-\frac{1}{2}\right)+\frac{\zeta \left(\frac{1}{2}\right)}{4}-\frac{\zeta \left(\frac{3}{2}\right)}{64}\right)+O\left(\left(\frac{1}{n}\right)\right)$$ which does not seem to be too bad $(46.0854$ for $n=10)$.

For $n=1000$, the value of the sum is $42210.32452$ while the approximation gives $42210.31462$.

Answer 2

There is no simple closed formula for such a sum, but it is not too difficult to guess that it behaves like: $$\sum_{n=1}^{N}\sqrt{4n+1}\approx \frac{1}{6}(4N+1)^{3/2}.$$ As a matter of fact, the inequality: $$ \sum_{n=1}^{N}\sqrt{4n+1}= \frac{1}{6}(4N+1)^{3/2}+\frac{1}{2}(4N+1)^{1/2}+\theta$$ where $\theta\in(-0.82,-0.745)$ holds by induction.