How can I solve the following summation?

from 0 to infinity

0.5^(n+1)/(2n+1)!

Answer 1

[Walter Roberson]

You use the Symbolic Toolbox, calling evalin() or feval() to use the toolbox's sum() command.

The answer is (1/4)*sqrt(2)*(exp((1/2)*sqrt(2))-exp(-(1/2)*sqrt(2))) which is also expressible as (1/2)*sqrt(2)*sinh((1/2)*sqrt(2))

For example,

syms n
feval(symengine, 'sum', 1/factorial(n), 'n = 0 .. infinity')

Answer 2

[Roger Stafford]

That's a convergent series, Kei Hin, but it would be a lot easier to evaluate sqrt(.5)*sinh(sqrt(.5)) which gives the same value.

Answer 3

[Image Analyst]

kei: You for got to do the sum in your code. Try this:

numberOfTerms = str2double(cell2mat(inputdlg('Enter the last term number')))
sumOfTerms = 0;
whos numberOfTerms
for n = 0 : numberOfTerms
  r = 0.5^(n+1)/factorial(2*n+1);
  sumOfTerms = sumOfTerms + r;
  fprintf('For term #%d, r = %.5f, and sumOfTerms = %.5f\n',...
    n, r, sumOfTerms);
end