Convergence of ∑n=1∞e−xnln⁡xx2+n if x&gt;0

I appear to get a different answer.an=x−nxx2+n⇒limn→∞(an)1n=limn→∞x−x(x2+n)1n=limn→∞x−xx2n(1+nx2)1nNow x2n→1 and (1+nx2)1n→ennx2=e1x2⇒limn→∞(an)1n=x−xe1x2Plugging this in a graphing calculator (or equivalently analyzing using derivatives) shows that its always lesser than 1. So, shouldn't the series converge .

Expert answers to "Convergence of ∑n=1∞e−xnln⁡xx2+n if x>0"

High School
Calculus and Analysis
Calculus 2
Series

Tajzler6ek

Answered question

2022-02-25

Convergence of

\sum_{n = 1}^{\infty} \frac{e^{- x n} \ln x}{x^{2} + n}

x > 0

Answer & Explanation

Lana Barlow

Beginner2022-02-26Added 4 answers

I appear to get a different answer.

a_{n} = \frac{x^{- n x}}{x^{2} + n}

\Rightarrow lim_{n \to \infty} {(a_{n})}^{\frac{1}{n}} = lim_{n \to \infty} \frac{x^{- x}}{{(x^{2} + n)}^{\frac{1}{n}}}

= lim_{n \to \infty} \frac{x^{- x}}{x^{\frac{2}{n}} {(1 + \frac{n}{x^{2}})}^{\frac{1}{n}}}

Now

x^{\frac{2}{n}} \to 1

and

{(1 + \frac{n}{x^{2}})}^{\frac{1}{n}} \to e^{\frac{n}{n x^{2}}} = e^{\frac{1}{x^{2}}}

\Rightarrow lim_{n \to \infty} {(a_{n})}^{\frac{1}{n}} = x^{- x} e^{\frac{1}{x^{2}}}

Plugging this in a graphing calculator (or equivalently analyzing using derivatives) shows that its always lesser than 1. So, shouldn't the series converge .

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