Calculate the integral part of $$\sum_{n=1}^{10^7} \frac{1}{\sqrt n}$$
and calculate $$\sum_{n=10^4}^{10^7} \frac{1}{\sqrt n}$$ to within an error of $\frac {1}{50}$.
I have found out that $\frac{1}{\sqrt n}$ is bounded by $2(\sqrt {n+1} - \sqrt{n})$ and $2(\sqrt {n} - \sqrt{n-1})$. This would mean $\sum_{n=1}^{10^7} \frac{1}{\sqrt n}$ is between $2(\sqrt {10^7+1} - \sqrt{1})$ and $2(\sqrt {10^7} - \sqrt{0})$. How would i find the second part and to within an error of $\frac{1}{50}$.
The second sum.
By your work and by the telescopic summation we have: $$6124.5556...<\Sigma<6124.5653...,$$ which says that $$\Sigma\approx6124.56$$ The telescoping summation gives: $$\sum_{n=10^4}^{10^7}\frac{1}{\sqrt{n}}<2\sum_{n=10^4}^{10^7}(\sqrt{n}-\sqrt{n-1})=2\left(\sqrt{10^7}-\sqrt{10^4-1}\right)=6124.5653...$$ and $$\sum_{n=10^4}^{10^7}\frac{1}{\sqrt{n}}>2\sum_{n=10^4}^{10^7}(\sqrt{n+1}-\sqrt{n})=2\left(\sqrt{10^7+1}-\sqrt{10^4}\right)=6124.5556...$$