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

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.

• Look at both sides of the inequality in the hint - you should be able to bound between $2(\sqrt{999}-\sqrt{1})$ and $2(\sqrt{1000}-\sqrt{2})$ (if I've done my algebra right, but regardless it will be something similar to this). Apr 11, 2018 at 18:25

It follows from $$\frac{1}{2\sqrt{k+1}} \le \sqrt{k+1} - \sqrt{k} \le \frac{1}{2\sqrt{k}}$$ that $$\sum_{k=1}^{9998}\frac{1}{2\sqrt{k+1}} = \sum_{k=2}^{9999}\frac{1}{2\sqrt{k}} \le \sqrt{9999}-1$$ and $$\sqrt{10000}-\sqrt{2} \le \sum_{k=2}^{9999}\frac{1}{2\sqrt{k}}.$$
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 $$197.1716 \le \sum_{k=2}^{9999}\frac{1}{\sqrt{k}} \le 197.9900.$$
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 $$E\left(\sum_{k=2}^{10^4-1}\frac{1}{\sqrt{k}}\right)=197$$
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.