Test the series for convergence or divergence.
sum (n=1, infinity) (n!)^n / n^3n
sum (n=1, infinity) 1 / n+n cos^2(8n)
Please show work/ explain. Thanks so much.
sum (n=1, infinity) (n!)^n / n^3n
sum (n=1, infinity) 1 / n+n cos^2(8n)
Please show work/ explain. Thanks so much.
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The first diverges by the root test. n³ < n! for all n ≥ 5.
The second has be stumped. It should diverge, but Wolfram Alpha has it converging. (It showed the first converging too but to an astronomically large number, on the order of 10^(70,000), so I'm guessing a calculation error.) It should diverge because 0 < cos²(8n) < 1 for all n ≥ 1, meaning the denominator is bounded by 2n which in turn means that 1/(2n) is less than the nth term for all n ≥ 1. Since 1/(2n) diverges so should the one shown above. Honestly, I'm a bit confused, but all my being is telling me that Wolfram is wrong.
If anyone else can explain why this would converge I'd be more than happy to read.
The second has be stumped. It should diverge, but Wolfram Alpha has it converging. (It showed the first converging too but to an astronomically large number, on the order of 10^(70,000), so I'm guessing a calculation error.) It should diverge because 0 < cos²(8n) < 1 for all n ≥ 1, meaning the denominator is bounded by 2n which in turn means that 1/(2n) is less than the nth term for all n ≥ 1. Since 1/(2n) diverges so should the one shown above. Honestly, I'm a bit confused, but all my being is telling me that Wolfram is wrong.
If anyone else can explain why this would converge I'd be more than happy to read.
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sum (n=1, infinity) (n!)^n / n^3n
Apply the Root Test
The nth root of the nth term is [(n!)^n / n^3n]^(1/n)
= n! / n^3
Since n^3 < n! for , as n approaches infinity, n!/n^3 approaches infinity.
Therefore, by Root Test, the series diverges.
Apply the Root Test
The nth root of the nth term is [(n!)^n / n^3n]^(1/n)
= n! / n^3
Since n^3 < n! for , as n approaches infinity, n!/n^3 approaches infinity.
Therefore, by Root Test, the series diverges.