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Sum of near infinite series?

Asked by John Mitchell on 22 Oct 2019 at 3:13
Latest activity Commented on by David Hill on 22 Oct 2019 at 21:22

Having some trouble figuring out where to start with this problem:
Find the product of the first n (n=9999) terms of the following expression where n is an integer variable greater than one:
eq = 2/1 * 2/3 * 4/3 * 4/5 * 6/5 * 6/7 * 8/7 * 8/9 .... *n-1/n
Any help on figuring this out would be appreciated.

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2 Answers

Answer by David Hill on 22 Oct 2019 at 3:36

p=[];
for i=1:2:n+1
p=[p,(n+1)/n,(n+1)/(n+2)];
end
P=prod(p(1:n));

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John Mitchell on 22 Oct 2019 at 3:54
Thank you for responding, but I forgot to mention that we can't use for loops in the problem, is there any way to do them without loops?
David Hill on 22 Oct 2019 at 21:22
a=2*[1:floor((n+1)/2);1:floor((n+1)/2)];
nu=a(:)';
d=[1,nu-1];
p=prod(nu(1:n)/d(1:n));

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Stephan
Answer by Stephan
on 22 Oct 2019 at 4:30

Think about your row. You can rewrite it - always two elements are the same as n^2/(n^2-1).
With this knowledge read about cumprod. This should be enough to bring you to work the rest by yourself.

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