Saturday

December 3, 2016
Posted by **maria PLEASE HELP** on Friday, March 9, 2012 at 2:40pm.

((-1)^(n+1)*n!)/(n*n) (from n=1 to infinity)

Within the MAIN function:

. Read a variable EPSILON of type DOUBLE (desired accuracy) from

the standard input.

EPSILON is an extremely small positive number which can be anything

between 1E-06 (10^(-6)) and 1E-12 (10^(-12)).

. EPSILON value will be passed to the FUNCTION as an argument.

Within the FUNCTION:

. In a do-while loop:

. Continue adding up the terms until |Sn+1 - Sn| < EPSILON.

. Sn is the sum of the first n-terms.

. Sn+1 is the sum of the first (n+1)-terms.

When the desired accuracy EPSILON is reached print the SUM and the number

of TERMS added to the sum.

- C++ Typo? -
**MathMate**, Friday, March 9, 2012 at 4:34pmYou may want to check the question.

If I interpret your question correctly, you are to calculate:

∞

Σ [(-1)^(i+1)]i!/i^2

i=1

is divergent.

You can show divergence by seeing each term (without regard to sign) as

T(i-1)=(i-1)!/(i-1)^2

So

T(i)=i!/i^2

=[T(i-1)*i]*(i-1)^2/i^2

=T(i-1) *(i-1)^2/i

which is clearly divergent. - C++ -
**maria PLEASE HELP**, Friday, March 9, 2012 at 5:02pmexcuse me if a run this in a c program. is it will be right or no?

- C++ -
**MathMate**, Friday, March 9, 2012 at 5:11pmIt will never converge, because the terms are getting bigger and bigger.

For the sum of a series to converge, the limit of the nth term must be zero, when n->∞.

The limit of n!/n^2 is ∞, so the series will never converge.

Take the example of the first few terms:

n=1, T(1)=1!/1^2=1

n=2, T(2)=-2!/2^2=1/2

n=3, T(3)=3!/3^2=2/3

n=4, T(4)=-4!/4^2=3/2

n=5, T(5)=5!/5^2=24/5

n=6, T(6)=-6!/6^2=20

....

You can continue with the computer and the calculator.

I suspect there is a typo. One possibility is that the expression is meant to be:

((-1)^(n+1)*n!)/(n**^**n) (from n=1 to infinity) - C++ -
**maria PLEASE HELP**, Friday, March 9, 2012 at 5:23pmAlso ý don't understand the question. this question was publishted in system this is my homework of c++. and question was exactly same which ý wrote up. ý don't know how ý should to download in system the homework in c program or normaly in text. what you think how ý should to do this. ý think question is just to compute the forward series. do you have any idea??

- C++ -
**MathMate**, Friday, March 9, 2012 at 5:30pmPlease confirm if the denominator was n^n or n*n, it makes the whole difference.

I do not think you should start working on C++ until you get the question cleared up. C++ is just a tool to get the answer, if there is one. - C++ -
**MathMate**, Friday, March 9, 2012 at 5:35pmBy the way, here are the first 40 sums according to your question. Hope it is obvious that it is not going to converge.

1 1.000000

2 0.500000

3 1.166667

4 -0.333333

5 4.466667

6 -15.533333

7 87.323810

8 -542.676190

9 3937.323810

10 -32350.676190

11 297540.232900

12 -3028859.767100

13 33817417.155977

14 -410969782.844023

15 5400916297.155977

16 -76328731702.844020

17 1154423614650.097200

18 -18605989057349.902000

19 318361048086229.060000

20 -5763893972355371.000000

21 110088582607484624.000000

22 -2212226970652035330.000000

23 46657369889243955000.000000

24 -1030509994230963400000.000000

25 23787426075098613000000.000000

26 -572797575591477840000000.000000

27 14363923206875396000000000.000000

28 -374524271450922870000000000.000000

29 10138866922056450000000000000.000000

30 -284586532869266930000000000000.000000

31 8271957332040122000000000000000.000000

32 -248691750611019940000000000000000.000000

33 7724969974652419000000000000000000.000000

34 -247666724696285400000000000000000000.000000

35 8187556105006690000000000000000000000.000000

36 -278844331850163950000000000000000000000.000000

37 9775029365174193000000000000000000000000.000000

38 -352428999351308500000000000000000000000000.000000

39 13058407345946156000000000000000000000000000.000000

40 -496888644683989900000000000000000000000000000.000000 - C++ -
**maria PLEASE HELP**, Friday, March 9, 2012 at 6:04pmý took this from web site copy paste.. ý am sure is like this

- C++ -
**MathMate**, Friday, March 9, 2012 at 6:38pmI suggest you talk to the teacher or the author of the website and explain that the series diverges.

If it is a public site, I'd like to know the link.

On the other hand, I don't know if you got the right web site, I believe your question probably started with:

"Write a MAIN function and a FUNCTION to compute the sum of the below series.

1 + 2!/4 + 3!/27 + 4!/256 + ...

Determine the general term of the series first.

......

......"

from which the general term is

n!/(n^n) (and not n!/n^2)

except that your series is an alternating series. - C++ -
**maria PLEASE HELP**, Saturday, March 10, 2012 at 5:29amQUESTÝON ÝS EXACTLY THÝS :

Write a MAIN function and a FUNCTION to compute the sum of the below series.

((-1)^(n+1)*n!)/(n*n) (from n=1 to infinity)

Within the MAIN function:

. Read a variable EPSILON of type DOUBLE (desired accuracy) from

the standard input.

EPSILON is an extremely small positive number which can be anything

between 1E-06 (10^(-6)) and 1E-12 (10^(-12)).

. EPSILON value will be passed to the FUNCTION as an argument.

Within the FUNCTION:

. In a do-while loop:

. Continue adding up the terms until |Sn+1 - Sn| < EPSILON.

. Sn is the sum of the first n-terms.

. Sn+1 is the sum of the first (n+1)-terms.

When the desired accuracy EPSILON is reached print the SUM and the number

of TERMS added to the sum.

TEST the program with different EPSILON values between 1E-06 and 1E-12. - C++ -
**MathMate**, Saturday, March 10, 2012 at 7:15amThere is a mistake in the question because the series will NOT converge, and will become an infinite loop (never stops).

The results are as shown above.

If you want to continue coding while waiting for your teacher's instructions (it's only one character to correct), you can proceed, but beware that it will run indefinitely if coded*correctly*.

Have you started coding? I do not think you need a pseudocode, since the instructions are explicit. - C++ -
**MathMate**, Saturday, March 10, 2012 at 10:31amFor your information, if the denominator had been n^n, the series converges to 0.655832 in about 17 iterations, and to 0.655831600867 (12 digits) in 34 iterations.

If the denominator remains n*n (= n^2), the results are shown to diverge, as demonstrated in the above post

(Friday, March 9, 2012 at 5:35pm)

There is also the possibility that the expression n*n was originally n**n which meant n^n in fortran (an earlier language). - C++ -
**maria PLEASE HELP**, Saturday, March 10, 2012 at 11:02amwhat is meaning of this which said: TEST the program with different EPSILON values between 1E-06 and 1E-12.

if ý send to teacher this answer:

You may want to check the question.

If I interpret your question correctly, you are to calculate:

∞

Σ [(-1)^(i+1)]i!/i^2

i=1

is divergent.

You can show divergence by seeing each term (without regard to sign) as

T(i-1)=(i-1)!/(i-1)^2

So

T(i)=i!/i^2

=[T(i-1)*i]*(i-1)^2/i^2

=T(i-1) *(i-1)^2/i

which is clearly divergent.

is goona be a right or not.. summary ý want to know which answer ý need to send :S:S - C++ -
**MathMate**, Saturday, March 10, 2012 at 7:26pmThe answer the teacher expects from you is the C++ code which will calculate the sum of the given series.

What you can do is to demonstrate using you code that the series diverges, unless the teacher modifies the question in the mean time.

EPSILON is a constant which will end the calculations when the difference between successive terms is less than EPSILON.

For EPSILON = 1E-6, the series ((-1)^(n+1)*n!)/(n^n) (from n=1 to infinity) will converge in 17 terms, and when EPSILON =1E-12, it will converge in 34 terms. So the value of EPSILON determines how accurate the sum would be.

For the given series (by your teacher), the series will not converge, because the difference between successive terms increase indefinitely. So it will become an infinite loop. - C -
**maria**, Tuesday, March 13, 2012 at 2:50pmyou are right the question is exactly like this... can you help me for this question.... thanks

Write a MAIN function and a FUNCTION to compute the sum of the below series.

((-1)^(n+1)*n!)/(n^n) (from n=1 to infinity)

Within the MAIN function:

. Read a variable EPSILON of type DOUBLE (desired accuracy) from

the standard input.

EPSILON is an extremely small positive number which can be anything

between 1E-06 (10^(-6)) and 1E-12 (10^(-12)).

. EPSILON value will be passed to the FUNCTION as an argument.

Within the FUNCTION:

. In a do-while loop:

. Continue adding up the terms until |Sn+1 - Sn| < EPSILON.

. Sn is the sum of the first n-terms.

. Sn+1 is the sum of the first (n+1)-terms.

When the desired accuracy EPSILON is reached print the SUM and the number

of TERMS added to the sum.