GRE Quantitative Reasoning - Sequences & Series

Evaluate: $\large \large \sum_{x=1}^{10} \frac {1}{n}$ . (Round to 4 places)

$\large \large \large 2.929$

$\large \large 2.9289$

$\large \large 2.92903$

$\large 2.92897$

Explanation

Step 1: Plug in values into the function and add up the fraction:

$\large 1+\frac {1}{2}+\frac {1}{3}+\frac {1}{4}+\frac {1}{5}+\frac {1}{6}+\frac {1}{7}+\frac {1}{8}+\frac {1}{9}+\frac {1}{10}$

Step 2: Find the sum of the fractions....

We can convert the fractions to decimals:

$\large 1+.5+.3333+.25+.2+.1666+.142857+.125+.1111+.1$

$\large =2.92896$

Step 3: Round to $\large 4$ places...

$\large \large 2.92896 \implies 2.9290\implies 2.929$

Find the radius of convergence for the power series

$\small \sum_{n=0}^{\infty}\frac{(n!)^2x^n}{(2n)!}$

$\small R=4$

$\small \small R=\frac{1}{4}$

$\small \small R=\infty$

$\small \small R=2$

$\small R=0$

Explanation

We can use the limit

$\small \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right|$

to find the radius of convergence. We have

$\small \small \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right|=\lim \left| \frac{x^{n+1}[(n+1)!]^2}{(2(n+1))!} \frac{(2n!)}{(n!)^2x^n}\right|$

$\small \small \small \small =\lim |x|\left| \frac{(2n)!}{(2n+2)!} \frac{[(n+1)!]^2}{(n!)^2}\right|=\lim |x|\left| \frac{(n+1)(n+1)}{(2n+1)(2n+2)}\right|=\frac{|x|}{4}<1$

$\small \implies |x|<4=R$

This means the radius of convergence is $\small R=4$ .

Evaluate: $\large \large \sum_{x=1}^{10} \frac {1}{n}$ . (Round to 4 places)

$\large \large \large 2.929$

$\large \large 2.9289$

$\large \large 2.92903$

$\large 2.92897$

Explanation

Step 1: Plug in values into the function and add up the fraction:

$\large 1+\frac {1}{2}+\frac {1}{3}+\frac {1}{4}+\frac {1}{5}+\frac {1}{6}+\frac {1}{7}+\frac {1}{8}+\frac {1}{9}+\frac {1}{10}$

Step 2: Find the sum of the fractions....

We can convert the fractions to decimals:

$\large 1+.5+.3333+.25+.2+.1666+.142857+.125+.1111+.1$

$\large =2.92896$

Step 3: Round to $\large 4$ places...

$\large \large 2.92896 \implies 2.9290\implies 2.929$

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

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

We cannot conclude when we use the ratio test.

It is convergent.

$\infty$

$-\infty$

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.

Find the radius of convergence for the power series

$\small \sum_{n=0}^{\infty}\frac{(n!)^2x^n}{(2n)!}$

$\small R=4$

$\small \small R=\frac{1}{4}$

$\small \small R=\infty$

$\small \small R=2$

$\small R=0$

Explanation

We can use the limit

$\small \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right|$

to find the radius of convergence. We have

$\small \small \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right|=\lim \left| \frac{x^{n+1}[(n+1)!]^2}{(2(n+1))!} \frac{(2n!)}{(n!)^2x^n}\right|$

$\small \small \small \small =\lim |x|\left| \frac{(2n)!}{(2n+2)!} \frac{[(n+1)!]^2}{(n!)^2}\right|=\lim |x|\left| \frac{(n+1)(n+1)}{(2n+1)(2n+2)}\right|=\frac{|x|}{4}<1$

$\small \implies |x|<4=R$

This means the radius of convergence is $\small R=4$ .

Evaluate: $\large \large \sum_{x=1}^{10} \frac {1}{n}$ . (Round to 4 places)

$\large \large \large 2.929$

$\large \large 2.9289$

$\large \large 2.92903$

$\large 2.92897$

Explanation

Step 1: Plug in values into the function and add up the fraction:

$\large 1+\frac {1}{2}+\frac {1}{3}+\frac {1}{4}+\frac {1}{5}+\frac {1}{6}+\frac {1}{7}+\frac {1}{8}+\frac {1}{9}+\frac {1}{10}$

Step 2: Find the sum of the fractions....

We can convert the fractions to decimals:

$\large 1+.5+.3333+.25+.2+.1666+.142857+.125+.1111+.1$

$\large =2.92896$

Step 3: Round to $\large 4$ places...

$\large \large 2.92896 \implies 2.9290\implies 2.929$

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

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

We cannot conclude when we use the ratio test.

It is convergent.

$\infty$

$-\infty$

Explanation

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.

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

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

We cannot conclude when we use the ratio test.

It is convergent.

$\infty$

$-\infty$

Explanation

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.

Find the radius of convergence for the power series

$\small \sum_{n=0}^{\infty}\frac{(n!)^2x^n}{(2n)!}$

$\small R=4$

$\small \small R=\frac{1}{4}$

$\small \small R=\infty$

$\small \small R=2$

$\small R=0$

Explanation

We can use the limit

$\small \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right|$

to find the radius of convergence. We have

$\small \small \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right|=\lim \left| \frac{x^{n+1}[(n+1)!]^2}{(2(n+1))!} \frac{(2n!)}{(n!)^2x^n}\right|$

$\small \small \small \small =\lim |x|\left| \frac{(2n)!}{(2n+2)!} \frac{[(n+1)!]^2}{(n!)^2}\right|=\lim |x|\left| \frac{(n+1)(n+1)}{(2n+1)(2n+2)}\right|=\frac{|x|}{4}<1$

$\small \implies |x|<4=R$

This means the radius of convergence is $\small R=4$ .

If the first term of an arithmetic sequence is 2 and the third term is 8, find the $\large \large 16$ th term.

$\large 47$

$\large 49$

$\large 46$

$\large 43$

Explanation

Step 1: Find the difference between each term...

Subtract the first term from the third term...

$\large 8-2=6$
There are two terms between first and third...Take the answer in step 1 and divide by 2 to get the difference between consecutive terms...