Let $p=1+\frac{1}{\sqrt 2}+\cdots\frac{1}{\sqrt {120}}$ and $q=\frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots\frac{1}{\sqrt {121}}$ then

Question

Let $$p=1+\frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots\frac{1}{\sqrt {120}}$$ and $$q=\frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\frac{1}{\sqrt 4}+\cdots\frac{1}{\sqrt {121}}$$ then which of the following is true:
A)$\ p>20$
B)$\ q<20$
C)$\ p+q<40$
D)$\ p+q>40$

The answer is ABD.

I was able to obtain A and B, but was unable to obtain D.

My Attempt:

Consider the function $f(x)=\frac{1}{\sqrt x}$ when $x\in [a,a+1]$, then we have: $$\frac{1}{\sqrt {a+1}}\leq\frac{1}{\sqrt x}\leq \frac{1}{\sqrt a}$$ Thus, $$\int_a^{a+1}\frac{dx}{\sqrt {a+1}}<\int_a^{a+1}\frac{dx}{\sqrt x}< \int_a^{a+1}\frac{dx}{\sqrt a}$$ Or, $$\frac{1}{\sqrt {a+1}}<\int_a^{a+1}\frac{dx}{\sqrt x}< \frac{1}{\sqrt a}$$ Summing the above inequality from $a=1$ to $a=120$ gives $$q<\int_1^{121} \frac{dx}{\sqrt x}<p$$ That simplifies to $q<20<p$. But that doesn't help me comment on $p+q$.

Any help would be awesome.

@YvesDaoust $p-q=\frac{10}{11}$, but I am still unable to see anything. Sorry for my short-sighted brain. — AryanSonwatikar, Aug 12 '20 at 10:21
@YvesDaoust (insert flushed face) I'm still unable to see anything. Terribly short-sighted, I am. — AryanSonwatikar, Aug 12 '20 at 10:26
May be useful: https://math.stackexchange.com/questions/509840/use-induction-to-prove-that-1-frac-1-sqrt2-frac-1-sqrt3 — VIVID, Aug 12 '20 at 10:30
@YvesDaoust Uhh, if we had $q=19.1, p-q=\frac{10}{11}$ then $p=20.0090909090\cdots$, which doesn't really fulfill $p+q>40$. — AryanSonwatikar, Aug 12 '20 at 10:39
Here as sqrt(x) is a decreasing function so p>I>q. Where I is the integral of sqrt(x) from 1 to 121. Also as thr derivative of sqrt(x) is increasing, so p-I>I-q — user600016, Aug 12 '20 at 11:35

score 6 · Accepted Answer · edited Aug 12 '20 at 11:32

6

Here is one way:

Consider trapeziums formed by the points: $$(a,0),(a,f(a)),(a+1,0),(a+1,f(a+1))$$ for $a=1,2,....120$, and note that the sum of areas of these trapeziums is greater than $\int_1^{121} f(x) dx$. Also note that the sum of areas of the trapeziums is nothing but $\frac{p+q}{2}$. And the conclusion follows.

edited Aug 12 '20 at 11:32

Gary

31,845

answered Aug 12 '20 at 10:42

LM2357

4,083

Ahh, I see now. Thanks! – AryanSonwatikar Aug 12 '20 at 10:55
I took the liberty of editing your answer so that it is more clear for future readers. Hope you don't mind. – AryanSonwatikar Aug 12 '20 at 11:08
I was planning to do that later...Thanks. – LM2357 Aug 12 '20 at 11:17

score 2 · Answer 2 · answered Aug 12 '20 at 10:29

2

Hint:
Utilize approximation $$S(n)=\sum_{k=1}^n\frac{1}{\sqrt k} \sim \frac{4n+1}{2\sqrt n}-\sqrt 2 $$

answered Aug 12 '20 at 10:29

Naren

3,432

Thank you for helping, but the approximation you quote is not in our syllabus. – AryanSonwatikar Aug 12 '20 at 10:35
He was probably using Euler-Maclaurin Summation. – Diger Aug 12 '20 at 12:25

Let $p=1+\frac{1}{\sqrt 2}+\cdots\frac{1}{\sqrt {120}}$ and $q=\frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots\frac{1}{\sqrt {121}}$ then

2 Answers2