Infinite series can be added and subtracted with each other.

\(\displaystyle \sum_{n=1}^{\infty}a_{n} + \sum_{n=1}^{\infty}b_{n} = \sum_{n=1}^{\infty}( a_{n} + b_{n})\)

Since the 2 series are convergent, the sum of the convergent infinite series is also convergent.

Note: The starting value, in this case n=1, must be the same before adding infinite series together.

This is a fundamental property of series.

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

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

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 

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

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

By the geometric test, this series is convergent.

Test the second series 

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

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

By the geometric test, this series is convergent.

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

We can use the limit

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

to find the radius of convergence. We have

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

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

\(\displaystyle \small \implies |x|< 4=R\)

This means the radius of convergence is \(\displaystyle \small R=4\).

In order to figure if 

\(\displaystyle \sum_{n=1}^{\infty} \frac{n!}{2^{n+1}}\)

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 \(\displaystyle \sum a_n\). We define,

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

Then if 

\(\displaystyle L< 1\), the series is absolutely convergent.

\(\displaystyle L>1\), the series is divergent.

\(\displaystyle L=1\), the series may be divergent, conditionally convergent, or absolutely convergent.

Now lets apply the ratio test to our problem.

Let  

\(\displaystyle a_n=\frac{n!}{2^{n+1}}\)

and

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

Now 

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

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

Now lets simplify this expression to 

\(\displaystyle =\lim_{n\rightarrow \infty}\left | \frac{n+1}{2} \right |\)

\(\displaystyle =\frac{1}{2}\lim_{n\rightarrow \infty}\left | n+1 \right |\)

\(\displaystyle =\frac{1}{2}*\infty=\infty\).

Since \(\displaystyle \infty>1\),

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

This is an infinite geometric series.

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

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

Solution:

Value of \(\displaystyle a_{1}\) can be found by setting \(\displaystyle k = 0\)

\(\displaystyle a_{1}=17\)

r is the value contained in the exponent

\(\displaystyle r=\frac{1}{6}\)

\(\displaystyle S=\frac{a_{1}}{1-r}\)

 \(\displaystyle S=\frac{17}{1-\frac{1}{6}} = \frac{17*6}{5}\)

\(\displaystyle S=\frac{102}{5}\)

This is an alternating series test.

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

Step 1:

An alternating series can be identified because terms in the series will “alternate” between + and –, because of \(\displaystyle (-1)^n\)

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

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

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

1.        \(\displaystyle \lim_{n \rightarrow \infty} b_{n} = 0\)
2.        {\(\displaystyle b_{n}\)} is a decreasing sequence, or in other words \(\displaystyle b_{n=1}' < 0\)

Solution:

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

1. \(\displaystyle \lim_{n \rightarrow \infty}b_{n}\)

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

2. {\(\displaystyle b_{n}\)} 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 \(\displaystyle b_{n+1} < \frac{1}{1000}\)

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

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

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

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

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

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

We can convert the fractions to decimals:

\(\displaystyle \large 1+.5+.3333+.25+.2+.1666+.142857+.125+.1111+.1\)

\(\displaystyle \large =2.92896\)

Step 3: Round to \(\displaystyle \large 4\) places...

\(\displaystyle \large \large 2.92896 \implies 2.9290\implies 2.929\)