Divergence/Convergence of the Infinite Series

Question

Hello!

I'm taking a Undergraduate Calculus course and I came up with this problem, which I'm not sure whether below method is the best one to solve this.

Your thoughts on this is highly appreciated.

Many thanks!

Question:

Using a suitable test determine whether the infinite series,

Sigma n=1 - infinity [ (e^(n+1))/((ln(n))^n) ] converges or diverges

The method I followed:

I considered the the nth term of the series a_n = [ (e^(n+1))/((ln(n))^n) ]

Then I considered the Root test to determine the convergence of divergence of this series, which derived;

lim n-->infinity |a_n|^(1/n) = lim n-->infinity [(e^(1/n))/(ln(n))]

Hence,

lim n-->infinity

|a_n|^(1/n) = 1/infinity -->0 < 1

Hence, by Root test, the above series converges.

Is this a correct method or is(are) their any other more suitable method(s) for determining the convergence/divergence of this series?

Answer 1

A loop program on programmable calculator indeed shows ∑_{{from n=2 to ∞}}[e⁽ⁿ⁺¹⁾ ÷ (ln n)ⁿ]

converging to 315.248614; this value first appears at n = 50 and returns ever after.

Note however that n must run from 2 to 50 since n=1 yields e⁽¹⁺¹⁾ ÷ (ln 1)¹

or e² ÷ 0¹ or e² ÷ 0, which attempts division by 0.

By the Root Test, one would write lim_(n→+∞)[e⁽ⁿ⁺¹⁾ ÷ (ln n)ⁿ]^1/n as

lim_(n→+∞) e^1/n[eⁿ ÷ (ln n)ⁿ]^1/n which translates to e⁰ • [e/(ln n)] or

1 • 0 or 0. The result of zero then indicates convergence.