I have the following series:
$C_N = C_{N-1} + \dfrac{1}{2N\sum\limits_{i=1}^{N}\frac{1}{\sqrt{i}}}$
Assuming $C_0 = 0$, the first few terms are as follows:
$C_N = 0+\dfrac{1}{2}+\dfrac{1}{4(1+\frac{1}{\sqrt{2}})}+\dfrac{1}{6(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}})} +\cdots + \dfrac{1}{2N\sum\limits_{i=1}^{N}\frac{1}{\sqrt{i}}}$
This series seems to be slightly smaller than $\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} +\cdots$ which we know diverges.
Is there a valid way to verfiy whether or not this series converges? If so, is there a reasonable estimate for its value?
FYI - After running this sum through 4000 iterations ($C_{4000}$), its value seems to be around 1.037058.
From $$\frac1{2\sqrt{i+1}}\le\frac1{\sqrt{i+1}+\sqrt{i}}=\sqrt{i+1}-\sqrt{i}\le\frac1{2\sqrt{i}},$$ we see that $$\sum^{n+1}_{i=2}\frac1{\sqrt{i}}\le2(\sqrt{n+1}-1)\le\sum^{n}_{i=1}\frac1{\sqrt{i}},$$ i.e. $$\sum^{n}_{i=1}\frac1{\sqrt{i}}=2\sqrt{n}+O(1).$$ This means that your sum converges, because $$\sum^\infty_{n=1}\frac1{n^{3/2}}$$ converges. One can use those inequalities (and the numerical value of $\zeta(3/2)$) to give rough estimates from above and below of the sum. Calculating it with higher precision is tricky, though.