Question 1: What can you conclude to the sum of the series based on the sequence listed as follows?

[MATH]P=\frac{110}{100+1}+\frac{110}{100+2}+\frac{110}{100+2}+\cdots+\frac{110}{100+10}[/MATH]

A. [MATH]P\gt 1[/MATH].

B. [MATH]P\gt 10[/MATH].

C. [MATH]P\gt 100[/MATH].

Answer:

As long as the least in the given decreasing sequence is greater than $1$, then all of the terms listed in the sequence are greater than $1$. And the least term falls on the the last term, i.e. [MATH]\frac{110}{100+10}=1[/MATH].

Therefore, all of them are greater than $1$. Since there are $10$ terms in the series, then their sum must be greater than $1\times 10=10$.

Question 2: You're asked to prove the RHS of the inequality below by rewriting [MATH]\frac{1}{10}[/MATH] as a sum of some numbers, which of the following option would you adopt?

[MATH]\frac{1}{11}\lt \frac{1}{100+1}+\frac{1}{100+2}+\cdots+\frac{1}{100+10}\le \frac{1}{10}[/MATH]

A. [MATH]\frac{1}{10}=\underbrace{\frac{1}{100}+\frac{1}{100}+\cdots+\frac{1}{100}}_{10 terms}[/MATH]

B. [MATH]\frac{1}{10}=\frac{1}{50}+\frac{1}{25}+\frac{1}{25}[/MATH]

C. [MATH]\frac{1}{10}=\frac{1}{60}+\frac{1}{48}+\frac{1}{48}+\frac{1}{48}+\frac{1}{48}[/MATH]

Answer:

Symmetry is what is crucial in determining the answer for this question. We have $10$ terms in the series, it's necessary to write [MATH]\frac{1}{10}[/MATH] as a sum of $10$ terms.

Thus A is the answer.

Question 3: Are you able in using the hints provided in the preceding questions to prove the RHS of the inequality

[MATH]\frac{1}{11}\lt \color{yellow}\bbox[5px,green]{\frac{1}{100+1}+\frac{1}{100+2}+\cdots+\frac{1}{100+10}\lt \frac{1}{10}}[/MATH]?

A. Yes.

B. No.

Answer:

We want to prove the RHS of the given inequality above, i.e.

[MATH]\frac{1}{100+1}+\frac{1}{100+2}+\cdots+\frac{1}{100+10}\lt \frac{1}{10}[/MATH]

When we've decided to rewrite [MATH]\frac{1}{10}[/MATH] as a sum of [MATH]\underbrace{\frac{1}{100}+\frac{1}{100}+\cdots+\frac{1}{100}}_{10 terms}[/MATH], we'll start to work on there:

[MATH]\frac{1}{100+1}+\frac{1}{100+2}+\cdots+\frac{1}{100+10}\lt \underbrace{\frac{1}{100}+\frac{1}{100}+\cdots+\frac{1}{100}}_{10 terms}[/MATH]

It's not hard to note that each of the term on the left is less than [MATH]\frac{1}{100}[/MATH]:

[MATH]\frac{1}{100+1}=\frac{1}{101}\lt \frac{1}{100}[/MATH]

[MATH]\frac{1}{100+2}=\frac{1}{102}\lt \frac{1}{100}[/MATH]

[MATH]\frac{1}{100+3}=\frac{1}{103}\lt \frac{1}{100}[/MATH]

[MATH]\,\,\,\,\,\,\,\,\vdots[/MATH]

[MATH]\frac{1}{100+10}=\frac{1}{110}\lt \frac{1}{100}[/MATH]

Therefore, we can conclude that

[MATH]\begin{align*}\frac{1}{100+1}+\frac{1}{100+2}+\cdots+\frac{1}{100+10}&\lt \frac{10}{100}\\&\lt \dfrac{1}{10}\end{align*}[/MATH]

Question 4: What algebraic modification would you apply to the fraction [MATH]\frac{1}{11}[/MATH] if you're to use the similar trick in proving the RHS of the inequality to prove the LHS of the inequality?

[MATH]\color{yellow}\bbox[5px,purple]{\frac{1}{11}\lt \frac{1}{100+1}+\frac{1}{100+2}+\cdots+\frac{1}{100+10}}\color{black}\lt \frac{1}{10}[/MATH]?

A. [MATH]\frac{1}{11}=\underbrace{\frac{1}{55}+\frac{1}{55}+\cdots+\frac{1}{55}}_{5 terms}[/MATH]

B. [MATH]\frac{1}{11}=\underbrace{\frac{1}{88}+\frac{1}{88}+\cdots+\frac{1}{88}}_{8 terms}[/MATH]

C. [MATH]\frac{1}{11}=\underbrace{\frac{1}{110}+\frac{1}{110}+\cdots+\frac{1}{110}}_{10 terms}[/MATH]

Answer:

Again, we have $10$ terms in [MATH]\frac{1}{100+1}+\frac{1}{100+2}+\cdots+\frac{1}{100+10}[/MATH], it's only appropriate to also create another $10$ terms from [MATH]\frac{1}{11}[/MATH], therefore we know C is the answer, and we need to verify that our choice works by working on proving it to be correct.

Note that

[MATH]\frac{1}{110}\lt \frac{1}{100+1}=\frac{1}{101}[/MATH]

[MATH]\frac{1}{110}\lt \frac{1}{100+2}=\frac{1}{102}[/MATH]

$\,\,\,\,\,\,\,\,\vdots$

[MATH]\frac{1}{110}= \frac{1}{100+10}=\frac{1}{110}[/MATH]

Adding them all up yields the desired result:

[MATH]\frac{1}{11}\lt \frac{1}{100+1}+\frac{1}{100+2}+\cdots+\frac{1}{100+10}[/MATH]

Question 5: Let [MATH]P=\frac{1}{100+1}+\frac{1}{100+2}+\cdots+\frac{1}{100+10}[/MATH] and that [MATH]\frac{1}{11}\lt P \lt \frac{1}{10}[/MATH], determine the value for $X,\,Y$ if [MATH]X\lt \frac{1}{P}\lt Y[/MATH].

A. [MATH]\left(X,\,Y\right)=\left(10,\,11\right)[/MATH]

B. [MATH]\left(X,\,Y\right)=\left(100,\,110\right)[/MATH]

C. [MATH]\left(X,\,Y\right)=\left(\frac{1}{11},\,\frac{1}{10}\right)[/MATH]

D. [MATH]\left(X,\,Y\right)=\left(\frac{1}{110},\,\frac{1}{100}\right)[/MATH]

Answer:

This is a breeze if you've a strong mathematics foundation as:

If [MATH]\frac{1}{11}\lt P \lt \frac{1}{10}[/MATH], then [MATH]10\lt \frac{1}{P} \lt 11[/MATH].

That is, if you invert both sides of the inequality, (that is, flip fractions upside down on both sides), you have got to reverse the direction of the inequality sign.

Therefore A is the answer.

Question 6: Can you prove the following inequality effectively using the trick you used in the previous problems?

Let [MATH]Q=\frac{1}{2017^2+1}+\frac{1}{2017^2+2}+\frac{1}{2017^2+3}+\cdots+\frac{1}{2017^2+2017}[/MATH]

[MATH]2017\lt \frac{1}{Q}\lt 2018[/MATH]

A. Yes.

B. No.

Answer:

You should be able to process the following procedures after you learned the tricks in the above questions:

[MATH]\frac{1}{2017^2+1}+\frac{1}{2017^2+2}+\frac{1}{2017^2+3}+\cdots+\frac{1}{2017^2+2017}[/MATH] is a decreasing series with a total number of terms of $2017$, and largest term is the first term, [MATH]\frac{1}{2017^2+1}[/MATH] and the smallest term is the last term, [MATH]\frac{1}{2017^2+2017}[/MATH].

If the smallest term, the last term [MATH]\left(\frac{1}{2017^2+2017}=\frac{1}{2018(2017)}\right)\ge \frac{1}{2018(2017)}[/MATH], then all of the terms will be greater than [MATH]\frac{1}{2018(2017)}[/MATH].

We then can make the partial conclusion that [MATH]Q=\frac{1}{2017^2+1}+\frac{1}{2017^2+2}+\frac{1}{2017^2+3}+\cdots+\frac{1}{2017^2+2017}\gt \frac{1}{2018(2017)}\times 2017=\frac{1}{2018}[/MATH].

By the same token, if the largest term, the first term in the series of Q, [MATH]\frac{1}{2017^2+1}[/MATH] is less than [MATH]\frac{1}{2017^2}[/MATH], which is true, then all of the remaining terms will be less than [MATH]\frac{1}{2017^2}[/MATH].

We can then make the another half conclusion,

[MATH]Q=\frac{1}{2017^2+1}+\frac{1}{2017^2+2}+\frac{1}{2017^2+3}+\cdots+\frac{1}{2017^2+2017}\lt \frac{1}{2017^2}\times 2017=\frac{1}{2017}[/MATH].

Therefore, we get:

[MATH]\frac{1}{2018}\lt Q\lt \frac{1}{2017}[/MATH]

And so

[MATH]2017\lt \frac{1}{Q}\lt 2018[/MATH] (Q.E.D).

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