Sum of roots reciprocal

First of all, let’s find an upper and lower bound for F(i)

1. Lower bound for F(i):

It can easily be shown that: $2(\sqrt{i + 1} - \sqrt{i}) < F(i)$

Proof:

$(\sqrt{i + 1} - \sqrt{i}) = \frac{(\sqrt{i + 1} - \sqrt{i}) X (\sqrt{i + 1} + \sqrt{i})}{\sqrt{i + 1} + \sqrt{i}} =$

$\frac{1}{\sqrt{i + 1} + \sqrt{i}}$ < $\frac{1}{\sqrt{i} + \sqrt{i}}$ = $\frac{1}{2 \sqrt{i}}$

Therefore :

$2(\sqrt{i + 1} - \sqrt{i}) < F(i) ... (1)$

2. Upper bound for F(i):

Also, It can easily be shown that: F(i) < $2(\sqrt{i} - \sqrt{i - 1})$

Proof:

$(\sqrt{i} - \sqrt{i - 1}) = \frac{(\sqrt{i} - \sqrt{i - 1}) X (\sqrt{i} + \sqrt{i - 1})}{(\sqrt{i} + \sqrt{i - 1})}$ =

$\frac{1}{(\sqrt{i} + \sqrt{i - 1}} > \frac{1}{(\sqrt{i} + \sqrt{i}})) = \frac{1}{2\sqrt{i}}$

Therefore :

$2(\sqrt{i} - \sqrt{i - 1}) > F(i) ... (2)$

Now we have: $\sum_{i=1}^{1,000,000} F(i) = 1 + 0.001 + \sum_{i=2}^{999,999} F(i)$ = 1.001 + S1 ... (3) )

$Where S1 = \sum_{i=2}^{999,999} F(i)$

In order to calculate this total, let’s find the sum of lower and upper bound of it.

Value of lower bound sum = $\sum_{i=2}^{999,999} 2(\sqrt{i + 1} - \sqrt{i}) = 2 [(\sqrt{3} - \sqrt{2}) + (\sqrt{4} - \sqrt{3}) + … (\sqrt{1000000} - \sqrt{999999})]$ $= 2 [(\sqrt{1000000} - \sqrt{2})] = 2 [1000 - 1.414] = 2 [998.586] = 1997.172$

Value of upper bound sum = $\sum_{i=2}^{999,999} 2(\sqrt{i} - \sqrt{i - 1}) = 2 [(\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + … (\sqrt{999999} - \sqrt{999998})]$ $= 2 [\sqrt{999999} - 1] = 2 [999.999 - 1] = 2 [998.999] = 1997.998$

From (1), (2) and (3) we have:

$1.001 + 1997.172 < \sum_{i=1}^{1,000,000} F(i) < 1.001 + 1997.998$

$1998.173 < \sum_{i=1}^{1,000,000} F(i) < 1998.999$

Therefore, the whole part of: $\sum_{i=1}^{1,000,000} F(i) = 1998$

Answer = $\boxed{1998}$

Comparing the area under the curve for the decreasing function $\frac{1}{\sqrt{x}}$ , we have for any positive integral $N$ $1+ \int_{1}^{N} \frac{1}{\sqrt{x}} dx > \sum_{i=1}^{N} \frac{1}{\sqrt{N}} > \int_{1}^{N+1} \frac{dx}{\sqrt{x}}$ This implies, $2\sqrt{N} -1 > \sum_{i=1}^{N} \frac{1}{\sqrt{N}} > 2 \sqrt{N+1} -2$ Putting $N=10^6$ , we get $1999 > \sum_{i=1}^{N} \frac{1}{\sqrt{N}} > 1998$ This gives us the desired result.