Limits of Sequences

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GRE Quantitative Reasoning › Limits of Sequences

Questions 1 - 8

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 $\small R=4$ .

Calculate the sum of the following infinite geometric series:

$\sum_{k=0}^{\infty}17\left(\frac{1}{6}\right)^k$

$\frac{102}{5}$

$\frac{101}{5}$

$\frac{104}{5}$

$\frac{103}{5}$

Explanation

This is an infinite geometric series.

The sum of an infinite geometric series can be calculated with the following formula,

$S = \frac{a_{1}}{1-r}$ , where $a_{1}$ is the first value of the summation, and r is the common ratio.

Solution:

Value of $a_{1}$ can be found by setting

$a_{1}=17$

r is the value contained in the exponent

$r=\frac{1}{6}$

$S=\frac{a_{1}}{1-r}$

$S=\frac{17}{1-\frac{1}{6}} = \frac{17*6}{5}$

$S=\frac{102}{5}$

You have a divergent series $\sum_{n=1}^{\infty} a_{n}$ , and you multiply it by a constant 10. Is the new series $\sum_{n=1}^{\infty}10* a_{n}$ convergent or divergent?

Divergent

Convergent

Inconclusive

Explanation

This is a fundamental property of series.

For any constant c, if $\sum_{n=1}^{\infty} a_{n}$ is convergent then $\sum_{n=1}^{\infty} c*a_{n}$ is convergent, and if $\sum_{n=1}^{\infty} a_{n}$ is divergent, $\sum_{n=1}^{\infty} c* a_{n}$ is divergent.

$\sum_{n=1}^{\infty} a_{n}$ is divergent in the question, and the constant c is 10 in this case, so $\sum_{n=1}^{\infty} 10*a_{n}$ is also divergent.

Determine how many terms need to be added to approximate the following series within $\frac{1}{1000}$ :

$\sum_{n=1}^{\infty} (-1)^n\frac{4}{(n!)^2}$

Explanation

This is an alternating series test.

In order to find the terms necessary to approximate the series within $\frac{1}{1000}$ first see if the series is convergent using the alternating series test. If the series converges, find n such that $b_{n+1} < \frac{1}{1000}$

Step 1:

An alternating series can be identified because terms in the series will “alternate” between + and –, because of

Note: Alternating Series Test can only show convergence. It cannot show divergence.

If the following 2 tests are true, the alternating series converges.

$a_{n} = (-1)^n * b_{n}$

$\lim_{n \rightarrow \infty} b_{n} = 0$
{ $b_{n}$ } is a decreasing sequence, or in other words $b_{n=1}' < 0$

Solution:

$b_{n}=\frac{4}{(n!)^2}$

1. $\lim_{n \rightarrow \infty}b_{n}$

$\lim_{n \rightarrow \infty}\frac{4}{(n!)^2} =0$

2. { $b_{n}$ } is a decreasing functon, since a factorial never decreases.

Since the 2 tests pass, this series is convergent.

Step 2:

Plug in n values until $b_{n+1} < \frac{1}{1000}$

$b_{4}=\frac{4}{(4!)^2}=\frac{1}{144} > \frac{1}{1000}$

$b_{5}=\frac{4}{(5!)^2}=\frac{1}{3600} < \frac{1}{1000}$

4 needs to be added to approximate the sum within $\frac{1}{1000}$ .

Determine if the following series is divergent, convergent or neither.

$\sum_{n=1}^{\infty} \frac{n!}{2^{n+1}}$

Divergent

Neither

Convergent

Both

Inconclusive

Explanation

In order to figure if

$\sum_{n=1}^{\infty} \frac{n!}{2^{n+1}}$

is convergent, divergent or neither, we need to use the ratio test.

Remember that the ratio test is as follows.

Suppose we have a series $\sum a_n$ . We define,

$L=\lim_{n\rightarrow \infty}\left | \frac{a_{n+1}}{a_n} \right |$

Then if

, the series is absolutely convergent.

, the series is divergent.

, the series may be divergent, conditionally convergent, or absolutely convergent.

Now lets apply the ratio test to our problem.

Let

$a_n=\frac{n!}{2^{n+1}}$

and

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

Now

$L=\lim_{n\rightarrow \infty}\left | \frac{\frac{(n+1)!}{2^{n+2}}}{\frac{n!}{2^{n+1}}} \right |$

$=\lim_{n\rightarrow \infty}\left | \frac{2^{n+1}(n+1)!}{2^{n+2}n!} \right |$ .

Now lets simplify this expression to

$=\lim_{n\rightarrow \infty}\left | \frac{n+1}{2} \right |$

$=\frac{1}{2}\lim_{n\rightarrow \infty}\left | n+1 \right |$

$=\frac{1}{2}*\infty=\infty$ .

Since $\infty>1$ ,

we have sufficient evidence to conclude that the series is divergent.

There are 2 series $\sum_{n=1}^{\infty} \left(\frac{3}{7}\right)^n$ and $\sum_{n=1}^{\infty}\left(\frac{5}{11}\right)^n$ .

Is the sum of these 2 infinite series convergent, divergent, or inconclusive?

Convergent

Divergent

Inconclusive

Explanation

A way to find out if the sum of the 2 infinite series is convergent or not is to find out whether the individual infinite series are convergent or not.

Test the first series

$\sum_{n=1}^{\infty} \left(\frac{3}{7}\right)^n$ .

This is a geometric series with $r = \frac{3}{7} < 1$ .

By the geometric test, this series is convergent.

Test the second series

$\sum_{n=1}^{\infty}\left(\frac{5}{11}\right)^n$ .

This is a geometric series with $r = \frac{5}{11} < 1$ .

By the geometric test, this series is convergent.

Since both of the series are convergent, $\sum_{n=1}^{\infty} \left(\frac{3}{7}\right)^n + \sum_{n=1}^{\infty}\left(\frac{5}{11}\right)^n$ is also convergent.

There are 2 series, $\sum_{n=1}^{\infty}a_{n}$ and $\sum_{n=1}^{\infty}b_{n}$ , and they are both convergent. Is $\sum_{n=1}^{\infty}a_{n} + \sum_{n=1}^{\infty}b_{n}$ convergent, divergent, or inconclusive?