The integer part of $ \sum_{k=2}^{9999} \frac{1}{\sqrt{k}} $

Question

For $k>0 $

$ \frac{1}{2\cdot \sqrt{k+1}}<\sqrt{k+1}-\sqrt{k}<\frac{1}{2\cdot \sqrt{k}} $,

the integer part of $ \sum_{k=2}^{9999} \frac{1}{\sqrt{k}} $

My attempt here was not complying with any of the options given as my attempt was to find the min. value for the series which the question does not ask. What to do?

Options are:

A. 198

B. 197

C. 196

D. 195

I know $ \sum_{k=1}^{n} \frac{1}{k} \approx log n $ But this does not help in this case.

Also $ \sum_{k=2}^{9999} \frac{1}{\sqrt{k}} < 2(\sqrt{10000}-\sqrt{2}) $

This also does not help.

Answer 1

It follows from $$ \frac{1}{2\sqrt{k+1}} \le \sqrt{k+1} - \sqrt{k} \le \frac{1}{2\sqrt{k}}$$ that \begin{equation} \sum_{k=1}^{9998}\frac{1}{2\sqrt{k+1}} = \sum_{k=2}^{9999}\frac{1}{2\sqrt{k}} \le \sqrt{9999}-1 \end{equation} and \begin{equation} \sqrt{10000}-\sqrt{2} \le \sum_{k=2}^{9999}\frac{1}{2\sqrt{k}}. \end{equation}

Thus we conlcude that $$ 98.5858 \approx \sqrt{10000}-\sqrt{2} \le \sum_{k=2}^{9999}\frac{1}{2\sqrt{k}} \le \sqrt{9999} - 1 \approx 98.9950 $$ which gives you the answer as $$ 197.1716 \le \sum_{k=2}^{9999}\frac{1}{\sqrt{k}} \le 197.9900.$$

Answer 2

Your inequality can be rewritten as: $$2 (\sqrt{k+1}-\sqrt{k})\leq \frac{1}{\sqrt{k}} \leq 2 (\sqrt{k}-\sqrt{k-1})$$ taking the sum: $$2 \sum_{k=2}^N (\sqrt{k+1}-\sqrt{k}) \leq \sum_{k=2}^N \frac{1}{\sqrt{k}} \leq \sum_{k=2}^N (\sqrt{k}-\sqrt{k-1})$$ i.e: $$2(\sqrt{N+1}-\sqrt{2}) \leq \sum_{k=2}^N \frac{1}{\sqrt{k}} \leq 2(\sqrt{N}-\sqrt{1})$$ you can then compute explicitly both terms.

Answer 3

for $t\in [k,k+1]$ we have, $$ \frac{1}{\sqrt{k+1}}\leq \frac{1}{\sqrt{t}}\leq \frac{1}{\sqrt{k}} $$ then $$ \frac{1}{\sqrt{k+1}}\leq \int_k^{k+1}\frac{dt}{\sqrt{t}}\leq \frac{1}{\sqrt{k}} $$ and we get \begin{eqnarray} \sum_{k=1}^{10^4-1}\frac{1}{\sqrt{k+1}}&\leq &\sum_{k=1}^{10^4-1}\int_k^{k+1}\frac{dt}{\sqrt{t}}&\leq& \sum_{k=1}^{10^4-1}\frac{1}{\sqrt{k}}\\ \sum_{k=2}^{10^4}\frac{1}{\sqrt{k}}&\leq &\int_1^{10^4}\frac{dt}{\sqrt{t}}&\leq& \sum_{k=1}^{10^4-1}\frac{1}{\sqrt{k}}\\ \frac{1}{100}+\sum_{k=2}^{10^4-1}\frac{1}{\sqrt{k}}&\leq &\int_1^{10^4}\frac{dt}{\sqrt{t}}&\leq& 1+\sum_{k=2}^{10^4-1}\frac{1}{\sqrt{k}}\\ \frac{1}{100}+\sum_{k=2}^{10^4-1}\frac{1}{\sqrt{k}}&\leq &2(100-1)&\leq& 1+\sum_{k=2}^{10^4-1}\frac{1}{\sqrt{k}}\\ \end{eqnarray} we can conclude that $$ 197\leq \sum_{k=2}^{10^4-1}\frac{1}{\sqrt{k}}\leq 198-\frac{1}{100}<198 $$ so $$ E\left(\sum_{k=2}^{10^4-1}\frac{1}{\sqrt{k}}\right)=197 $$