What diverges?

Calculus Level 3

Which of the following series diverges?

A: n Z + n 4 1 4 4 n + log 4 \displaystyle \sum\limits_{n\in\mathbb{Z^+}} \frac{n^4-\frac{1}{4}}{4^n+\log4}

B: n Z + ( n 2 + 3 n 4 n 5 + 6 ) n \displaystyle \sum\limits_{n\in\mathbb{Z^+}} (\frac{n^2+3n}{4n^5+6})^n

C: n Z + 5 sin ( n ) + 12 cos ( n ) 13 e n \displaystyle \sum\limits_{n\in\mathbb{Z^+}} \frac{5\sin(n)+12\cos(n)}{13e^n}

D: n Z + e ( n ! ) e n n e \displaystyle \sum\limits_{n\in\mathbb{Z^+}} \frac{e(n!)}{e^n-n^e}

C A D B

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2 solutions

J D
Jul 7, 2016

A factorial grows faster than an exponential, so the terms in series D do not tend to zero as n tends to infinity.

For the First option:-

Apply Ratio Test

lim n a n + 1 a n = 1 4 \lim_{n\to\infty}\frac{a_{n+1}}{a_{n}} = \frac{1}{4} So it converges.

For 2nd option :-

Applying Cauchy's nth root test we get the limit as 0 0 . Hence it converges

For 3rd option :-

We see that 5 s i n ( n ) + 12 c o s ( n ) 5sin(n)+12cos(n) is bounded . So ( 5 s i n ( n ) + 12 c o s ( n ) ) 1 n 1 , n (5sin(n)+12cos(n))^{\frac{1}{n}} \to 1 , n \to \infty

So again applying Cauchy's root test wer get the limit as 1 e \frac{1}{e} which is 1 \leq 1

For the 4th option:-

We check whether the nth term \to 0 0 as n \to \infty . 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 \infty and hence is not convergent .

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