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how do i fix this code? (finding convergence of divergence)

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Ann Lee
Ann Lee el 10 de Abr. de 2022
Comentada: Torsten el 10 de Abr. de 2022
hello!
cube=@(n)(factorial(n))/(n.^n);
this series is divergence over n being over 170.
and i want to make this series being converges when n being over 1000.
to make converges,
i want to conversion of expression to 'exp(ln(f(x))) = f(x)' (then it will work n being over 170..)
so i want to print when n= 170~1000, this series( 'exp(ln(f(n))) = f(n)' )'s sum and
find the result of sum is converges? or divergence?
and print sum of series's graph
how can i fix this code?
I would appreciate it if you could tell me what I can type in the commander window to see the graph.
function A=work5(x)
m=1;
y = 170:50:1000;
for k = y
n=1:k;
f(n)= sum(exp(ln(f(n))));
A(1,m)=f(n);
m=m+1;
end
fprintf('k-sum\t\t%-10s\n','series a)','C or D');
fprintf('%-d-sum\t\t%13.10f\t%13.10f\t%13.10f\n',[y; A]);
end

Respuesta aceptada

Torsten
Torsten el 10 de Abr. de 2022
cube = @(n)prod((1:n)/n);
  2 comentarios
Ann Lee
Ann Lee el 10 de Abr. de 2022
what should i change my code to 'cube = @(n)prod((1:n)/n);'?
Torsten
Torsten el 10 de Abr. de 2022
I don't understand what you mean.
If you want to calculate n! / n^n for a certain value of n, you can do
result = cube(n)
if you defined
cube = @(n) prod((1:n)/n);
before.

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