Which of the following series diverges?
A: n ∈ Z + ∑ 4 n + lo g 4 n 4 − 4 1
B: n ∈ Z + ∑ ( 4 n 5 + 6 n 2 + 3 n ) n
C: n ∈ Z + ∑ 1 3 e n 5 sin ( n ) + 1 2 cos ( n )
D: n ∈ Z + ∑ e n − n e e ( n ! )
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For the First option:-
Apply Ratio Test
lim n → ∞ a n a n + 1 = 4 1 So it converges.
For 2nd option :-
Applying Cauchy's nth root test we get the limit as 0 . Hence it converges
For 3rd option :-
We see that 5 s i n ( n ) + 1 2 c o s ( n ) is bounded . So ( 5 s i n ( n ) + 1 2 c o s ( n ) ) n 1 → 1 , n → ∞
So again applying Cauchy's root test wer get the limit as e 1 which is ≤ 1
For the 4th option:-
We check whether the nth term → 0 as n → ∞ . which is a necessary but not a sufficient condition for convergence of a series.
So applying Stirling's Approximation we see that
the limit tends to ∞ and hence is not convergent .
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A factorial grows faster than an exponential, so the terms in series D do not tend to zero as n tends to infinity.