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GRE Subject Test: Math : GRE Subject Test: Math

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Example Questions

Example Question #1 : Convergence And Divergence

Which of these series cannot be tested for convergence/divergence properly using the ratio test? (Which of these series fails the ratio test?)

Possible Answers:

$\sum_{k=0}^{\infty} \frac{(-2)^k}{k!}$

$\sum_{k=0}^{\infty} \frac{k^2}{3^k}$

$\sum_{k=0}^{\infty} \frac{1}{k!}$

None of the other answers.

$\sum_{k=0}^{\infty} k$

Correct answer:

$\sum_{k=0}^{\infty} k$

Explanation:

The ratio test fails when $\lim_{k\to \infty} \left |\frac{a_{k+1}}{a_k} \right | =1$ . Otherwise the series converges absolutely if $\lim_{k\to \infty} \left |\frac{a_{k+1}}{a_k} \right | <1$ , and diverges if $\lim_{k\to \infty} \left |\frac{a_{k+1}}{a_k} \right | >1$ .

Testing the series $\sum_{k=0}^{\infty} k$ , we have

$\lim_{k\to\infty} \left |\frac{a_{k+1}}{a_k} \right | = \lim_{k\to\infty} \left |\frac{(k+1)}{(k)} \right | = \lim_{k\to\infty} 1+ \frac{1}{k} = 1.$

Hence the ratio test fails here. (It is likely obvious to the reader that this series diverges already. However, we must remember that all intuition in mathematics requires rigorous justification. We are attempting that here.)

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Example Question #1 : Ratio Test

Assuming that $a_{n}=n^{\alpha}(n+1)$ , $\alpha \ge 2$ . Using the ratio test, what can we say about the series:

$\sum_{n=1}^{\infty} \frac{1}{a_{n}}$

Possible Answers:

We cannot conclude when we use the ratio test.

$-\infty$

It is convergent.

$\infty$

Correct answer:

We cannot conclude when we use the ratio test.

Explanation:

As required by this question we will have to use the ratio test. $L=\lim_{n\rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|$ if L<1 the series converges absolutely, L>1 the series diverges, and if L=1 the series could either converge or diverge.

To do so, we will need to compute : $\frac{a_{n+1}}{a_{n}}$ . In our case:

$a_{n}=\frac{1}{a_{n}}$

Therefore

$\frac{a_{n+1}}{a_{n}}=(\frac{n}{n+1})^{\alpha}\frac{n+1}{n+2}$ .

We know that $\lim_{n\rightarrow \infty }\frac{n}{n+1}=1$

This means that

$\frac{a_{n+1}}{a_{n}}=(\frac{n}{n+1})^{\alpha}\frac{n+1}{n+2}=1\forall\quad \alpha$

Since L=1 by the ratio test, we can't conclude about the convergence of the series.

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Example Question #2 : Ratio Test

We consider the series : $\sum_{n=1}^{\infty}\frac{1}{72n+89}$ , use the ratio test to determine the type of convergence of the series.

Possible Answers:

The series is fast convergent.

$\frac{1}{79}$

It is clearly divergent.

$\frac{3}{79}$

We cannot conclude about the nature of the series.

Correct answer:

We cannot conclude about the nature of the series.

Explanation:

To be able to use to conclude using the ratio test, we will need to first compute the ratio then use $L=\lim_{n\rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|$ if L<1 the series converges absolutely, L>1 the series diverges, and if L=1 the series could either converge or diverge. Computing the ratio we get,

$\frac{a_{n+1}}{a_{n}}$ .

We have then:

$a_{n+1}=\frac{1}{72(n+1)+89}=\frac{1}{72n+161}$

Therefore have :

$\frac{a_{n+1}}{a_{n}}=$ $\frac{72n+89}{72n+161}$

It is clear that $\lim_{n\rightarrow \infty}\frac{a_{n+1}}{a_{n}}=1$ .

By the ratio test , we can't conclude about the nature of the series.

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Example Question #2822 : Calculus Ii

Consider the following series :

$\sum_{n=1}^{\infty}a_{n}$ where $a_{n}$ is given by:

$a_{n}=\frac{n}{n^\alpha +1},\alpha \ge 3$ . Using the ratio test, find the nature of the series.

Possible Answers:

We can't conclude when using the ratio test.

$\frac{\pi}{e}$

$\frac{3}{211}$

The series is convergent.

Correct answer:

We can't conclude when using the ratio test.

Explanation:

Let $a_{n}$ be the general term of the series. We will use the ratio test to check the convergence of the series.

$L=\lim_{n\rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|$ if L<1 the series converges absolutely, L>1 the series diverges, and if L=1 the series could either converge or diverge.

We need to evaluate,

$\frac{a_{n+1}}{a_{n}}$ we have:

$a_{n+1}= \frac{n+1}{(n+1)^\alpha +1},a_{n}=\frac{n}{n^\alpha +1}$ .

Therefore:

$\frac{a_{n+1}}{a_{n}}=\frac{n^\alpha +1}{(n+1)^\alpha +1}\frac{n+1}{n}$ . We know that,

$\lim_{n\rightarrow \infty}\frac{n}{n+1}=1$ and therefore

$\lim_{n\rightarrow \infty}\frac{n^\alpha +1}{(n+1)^\alpha +1}=1$

This means that :

$\lim_{n\rightarrow \infty}\frac{a_{n}}{a_{n+1}}=1$ .

By the ratio test we can't conclude about the nature of the series. We will have to use another test.

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Example Question #1 : Polynomials

What are the roots of the polynomial: f(x)=x^2-7x-44 ?

Possible Answers:

x=-11, x=4

x=11,-4

x=-11,x=-4

None of the Above

Correct answer:

x=11,-4

Explanation:

Step 1: Find factors of 44:

(1,44), (2,22), (4,11)

Step 2: Find which pair of factors can give me the middle number. We will choose (4,11) .

Step 3: Using and , we need to get . The only way to get is if I have and .

Step 4: Write the factored form of that trinomial:

(x-11)(x+4)

Step 5: To solve for x, you set each parentheses to :

$(x-11)=0\rightarrow x=11$
$(x+4)=0\rightarrow x=-4$

The solutions to this equation are x=11 and x=-4 .

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Example Question #2 : Roots Of Polynomials

Solve for :

x^3-4x^2+x-4

Possible Answers:

$x=\pm i,4$

$x=\pm1,-4$

x=-1,4

x=1,4

Correct answer:

$x=\pm i,4$

Explanation:

Step 1: Factor by pairs:

x^3-4x^2+x-4=0

x^2(x-4)+1(x-4)=0

Step 2: Re-write the factorization:

(x^2+1)(x-4)=0

Step 3: Solve for x:

(x^2+1)=0;(x-4)=0
$\\ x^2=-1;x=4 \\\sqrt{x^2}=\pm\sqrt{-1}; x=4 \\x=\pm i;x=4$

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Example Question #1 : Roots Of Polynomials

Find :

x^2+5x-6

Possible Answers:

x=6

No Solutions Exist

x=1

x=1,-6

Correct answer:

x=1,-6

Explanation:

Step 1: Find two numbers that multiply to and add to .

We will choose 6,-1 .

Step 2: Factor using the numbers we chose:

(x+6)(x-1)=0

Step 3: Solve each parentheses for each value of x..

$(x+6)=0 \implies x=-6$
$(x-1)=0\implies x=1$

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Example Question #3 : Roots Of Polynomials

$Find\ all\ of\ the\ roots\ for\ the\ below\ polynomial:$

x^3+5x^2-3x-15=0

Possible Answers:

$x=\pm \sqrt{3}\ and\ -5$

$x= \sqrt{3}\ and\ -5$

$x=3\ and\ -15$

$x=3\ and\ 15$

Correct answer:

$x=\pm \sqrt{3}\ and\ -5$

Explanation:

Based upon the fundamental theorem of algebra, we know that there must exist 3 roots for this polynomial based upon its' degree of 3.

To solve for the roots, we use factor by grouping:

x^3+5x^2-3x-15=0

First group the terms into two binomials:

(x^3+5x^2)(-3x-15)=0

Then take out the greatest common factor from each group:

x^2(x+5)-3(x+5)=0

Now we see that the leftover binomial is the greatest common factor itself:

$(x+5)\ is\ common\ to\ both\ terms\ so\ we\ can\ take\ it\ out:$

(x+5)(x^2-3)=0

We set each binomial equal to zero and solve:

$x+5=0\ so\ x=-5$

$x^2-3=0;\ so\ x^2=3\ which\ means:$

$x=\pm\sqrt{3}\ after\ taking\ the\ square\ root\ of\ both\ sides.$

$The\ 3\ roots\ are\ -5, {\sqrt{3}},\ and\ -\sqrt{3}$

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Example Question #1 : Classifying Algebraic Functions

Find all of the roots for the polynomial below:

f(x)=x^3+21-7x^2-3x

Possible Answers:

$7,\ \sqrt{3}\ and\ -\sqrt{3}$

$7,\ and\ \sqrt{3}$

$3\ and\ \frac{3}{7}$

$-21,\ 3\ and\ \frac{3}{7}$

Correct answer:

$7,\ \sqrt{3}\ and\ -\sqrt{3}$

Explanation:

In order to find the roots for the polynomial we must first put it in Standard Form by decreasing exponent:

f(x)=x^3+21-7x^2-3x

x^3-7x^2-3x+21

Now we can use factor by grouping, we start by grouping the 4 terms into 2 binomials:

(x^3-7x^2)(-3x+21)

We now take the greatest common factor out of each binomial:

x^2(x-7)-3(x-7)

We can see that each term now has the same binomial as a common factor, so we simplify to get:

(x-7)(x^2-3)

To find all of the roots, we set each factor equal to zero and solve:

$x-7=0\ which\ gives\ us\ x=7$

$x^2-3=0\ which\ gives\ us\ x^2=3\ which\ then\ yields\ x=\pm \sqrt{3}$

$The\ roots\ are\ 7,\ \sqrt{3}\ and\ -\sqrt{3}$

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Example Question #5 : Roots Of Polynomials

$How\ many\ total\ roots\ will\ a\ polynomial\ with\ the\ below\ roots\ have?$

$x=-\sqrt{5};\ x=7; and\ x=3+2i$

Possible Answers:

Correct answer:

Explanation:

$If\ the\ complex\ root\ a+bi\ exists, then\ a-bi\ must\ also\ exist\ as\ a\ root.$

$If\ the\ irrational\ root\ a+\sqrt{b}\ exists,\ then\ the\ root\ a-\sqrt{b}\ must\ also\ exist.$

$Given\ that\ the\ above\ statements\ always\ hold\ true\ we\ know\ that:$

$If\ -\sqrt{5}\ exists\ as\ a\ root\, then\ +\sqrt{5}\ also\ exists\ as\ a\ root.$

$Also,\ if\ 3+2i\ exists\ as\ a\ root,\ then\ 3-2i\ must\ also\ exist\ as\ a\ root.$

$This\ gives\ us\ an\ additional\ 2\ roots\ above\ the\ 3\ roots\ we\ already\ know.$

$There\ will\ be\ a\ total\ of\ 5\ roots.$

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GRE Subject Test: Math : GRE Subject Test: Math

Study concepts, example questions & explanations for GRE Subject Test: Math

All GRE Subject Test: Math Resources

Example Questions

Example Question #1 : Convergence And Divergence

Example Question #1 : Ratio Test

Example Question #2 : Ratio Test

Example Question #2822 : Calculus Ii

Example Question #1 : Polynomials

Example Question #2 : Roots Of Polynomials

Example Question #1 : Roots Of Polynomials

Example Question #3 : Roots Of Polynomials

Example Question #1 : Classifying Algebraic Functions

Example Question #5 : Roots Of Polynomials

All GRE Subject Test: Math Resources

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