what is integer part of $$\sum_{j=2}^{100} \frac{1}{\sqrt{j}}= \frac{1}{\sqrt2}+\frac{1}{\sqrt3}+\frac{1}{\sqrt4}+\cdots+\frac{1}{\sqrt{100}}$$
I don’t have any idea how to approach this problem.
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3Here it is an idea: for any $n\geq 2$, $\frac{1}{\sqrt{n}}$ is pretty close to $$ 2\left(\sqrt{n+1/2}-\sqrt{n-1/2}\right),$$ which is a telescopic term. Draw your conclusions. You just have to resolve a small uncertainty between 17 and 18. – Jack D'Aurizio Apr 18 '18 at 11:52
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@JackD'Aurizio I read your answer but I still don’t get it ☹️ – Fawad Apr 18 '18 at 11:57
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What is that you don't get? Maybe $$\sum_{n=a}^{b}\left( f(n+1)-f(n) \right) = f(b+1)-f(a) $$ ? – Jack D'Aurizio Apr 18 '18 at 13:13
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1Contest math? Now we need to know the source. For questions from running contests are forbidden. – Jyrki Lahtonen Apr 18 '18 at 18:01
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@JyrkiLahtonen it was asked today in IIIT-H undergrad entrance exam. Test is over btw. – Fawad Apr 18 '18 at 18:02
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Thanks, Fawad. That's ok! – Jyrki Lahtonen Apr 18 '18 at 18:03
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2Still, why are we supporting questions without any attempt from OP? Voted to close as off-topic. – Jaideep Khare Apr 18 '18 at 18:08
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2This isn't rude behavior. This is what should be done. This isn't personal, but these type of question promote other users to ask homework questions. – Jaideep Khare Apr 18 '18 at 18:12
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yeah isn't this forum suppose students who can't solve a problem but tried? Why are you asking without attempt？ – Guy who failed everything Apr 20 '18 at 03:57
2 Answers
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Hint. One may use, for $n\ge1$, $$ 2\sqrt{n+1}-2\sqrt{n}= \int_{n}^{n+1}\frac{dx}{\sqrt{x}} < \frac{1}{\sqrt{n}}<\int_{n-1}^{n}\frac{dx}{\sqrt{x}}=2\sqrt{n}-2\sqrt{n-1}. $$ Then one may write $$ \sum_{j=2}^{100} \frac{1}{\sqrt{j}}=\frac1{\sqrt{2}}+\sum_{j=3}^{100} \frac{1}{\sqrt{j}} $$ and conclude using the above inequality for $n=3$ to $n=100$.
Olivier Oloa
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Here is a way by bounding the sum by integrals:
$$S = \sum_{j=2}^{100} \frac{1}{\sqrt{j}}= \frac{1}{\sqrt2}+\frac{1}{\sqrt3}+\frac{1}{\sqrt4}+\cdots+\frac{1}{\sqrt{100}} = 10 \sum_{j=2}^{100} \frac{1}{\sqrt{\frac{j}{100}}}\frac{1}{100}$$ Now, you can bound the Riemann-sum on the right by two integrals:
$$10\int_{\frac{2}{100}}^{\frac{101}{100}}\frac{1}{\sqrt{x}}dx < S < 10\int_{\frac{1}{100}}^{1}\frac{1}{\sqrt{x}}dx$$ You get $$17 < 2(\sqrt{101}-\sqrt{2}) <S < 2(\sqrt{100}-1) = 18 \Rightarrow [S] = 17$$
trancelocation
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1Can I know how you bounded that series using Rimann-sum. I only know when $\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{i=n}f(\frac{i}{n})=\int_0^1f(x)dx$ here $x=\frac{i}{n}$ thanks. – Fawad Apr 20 '18 at 04:55
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Just saw your question. Will answer later as I am travelling. Just as a quick help: The sim can be seen as an upper Darboux-sum for the left integral and as a lower Darboux-sum for the right integral. Note, that the interval of integration is slightly shifted for that reason. And note that the integrand is a postive decreasing function on the considered intervals. – trancelocation Apr 20 '18 at 05:17
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That “darboux sum” is not my level stuff, I will read about it once I am free,I have exam in few days. – Fawad Apr 20 '18 at 05:30
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1These are special Rieman-sums where you take instead of an arbitrary $f(\xi_i)$ for $\xi_i \in [x_i,x_{i+1}]$ always the minimum (lower Darboux-sum) or the maximum (upper Darboux-sum) of $f$ of the function on that small interval. – trancelocation Apr 20 '18 at 05:50